Free math worksheets

- Identify the three signed coefficients from standard form. - Substitute them into \(b^2-4ac\). - Check the sign of the constant coefficient carefully when multiplying.

1. Here \(a=2\), \(b=3\), and \(c=-4\). 2. \(D=b^2-4ac=3^2-4\cdot2\cdot(-4)=41\).

\(41\)

- Connect a vertex on the x-axis with a repeated x-intercept. - Compute \(b^2-4ac\) using the exact decimal coefficients shown. - State the discriminant value before interpreting its sign.

1. A vertex lies on the x-axis exactly when the quadratic has a repeated real zero, so \(D=0\). 2. a) \(D=7^2-4\cdot1\cdot12.25=0\), so the vertex lies on the x-axis. 3. b) \(D=(-2.4)^2-4\cdot1\cdot1.44=0\), so the vertex lies on the x-axis. 4. c) \(D=1^2-4\cdot1\cdot1=-3\), so the vertex does not lie on the x-axis.

a) \(D=0\); yes b) \(D=0\); yes c) \(D=-3\); no

- Keep the coefficients as fractions rather than converting to decimals. - Compare each discriminant with zero only after simplifying it exactly. - A negative discriminant means there are no real x-intercepts.

1. Touching the x-axis once requires \(D=0\). 2. a) \(D=\left(\frac23\right)^2-4\cdot1\cdot\frac19=0\). 3. b) \(D=\left(-\frac45\right)^2-4\cdot1\cdot\frac4{25}=0\). 4. c) \(D=\left(\frac12\right)^2-4\cdot1\cdot\frac12=-\frac74\).

a) \(D=0\); touches the x-axis once b) \(D=0\); touches the x-axis once c) \(D=-\frac74\); does not touch the x-axis and has no real x-intercepts

- A vertex on the x-axis corresponds to exactly one distinct real zero. - Write the coefficient triple for each function before calculating \(D\). - Distinguish a zero discriminant from a positive one.

1. a) \(D=10^2-4\cdot1\cdot25=0\), so the vertex is on the x-axis. 2. b) \(D=(-4)^2-4\cdot1\cdot3=4\), so there are two x-intercepts and the vertex is not on the x-axis. 3. c) \(D=5^2-4\cdot1\cdot6.25=0\), so the vertex is on the x-axis.

a) \(D=0\); yes b) \(D=4\); no c) \(D=0\); yes

- Compute the same discriminant expression for all three functions. - Distinguish exactly zero from merely nonnegative. - A positive discriminant corresponds to two distinct x-intercepts.

1. a) \(D=144-144=0\), so \(f\) touches once. 2. b) \(D=49-48=1\), so \(g\) crosses twice. 3. c) \(D=1-1=0\), so \(h\) touches once.

a) \(D=0\); touches once b) \(D=1\); does not touch once—it crosses twice c) \(D=0\); touches once

- Record \(a\), \(b\), and \(c\) with their signs before computing. - Only the sign of \(D\) is needed to classify the zeros. - Do not switch to factoring or the quadratic formula, because the roots are not requested.

1. a) \(D=36-36=0\), so there is one real zero. 2. b) \(D=4-10=-6\), so there are no real zeros. 3. c) \(D=16-4=12\), so there are two real zeros.

a) \(D=0\); one real zero b) \(D=-6\); no real zeros c) \(D=12\); two real zeros

- In parts a) and b), calculate \(b^2-4ac\) directly. - In part c), the two pieces of the discriminant are already supplied. - Part d) requires only interpreting a positive discriminant.

1. a) \(D=100-100=0\), so there is one real solution. 2. b) \(D=16-40=-24\), so there are no real solutions. 3. c) \(D=49-50=-1\), so there are no real solutions. 4. d) \(D=12.5>0\), so there are two distinct real solutions.

a) \(D=0\); one real solution b) \(D=-24\); no real solutions c) \(D=-1\); no real solutions d) \(D=12.5\); two distinct real solutions

- Identify the three coefficients. - Substitute them into \(b^2 - 4ac\). - Interpret the sign of the result.

1. The coefficients are \(a = \frac{1}{2}\), \(b = 3\), and \(c = 5\). 2. The discriminant is \(D = 3^2 - 4 \cdot \frac{1}{2} \cdot 5 = 9 - 10 = -1\). 3. Since \(D < 0\), the equation has no real solutions.

The discriminant is \(-1\), so there are no real solutions.

- Count the x-intercepts of each graph. - Relate the number of real zeros to the sign of the discriminant.

1. The graph of \(f\) crosses the x-axis twice, so \(f(x) = 0\) has two distinct real solutions. Therefore, \(D > 0\). 2. The graph of \(g\) has no x-intercepts, so \(g(x) = 0\) has no real solutions. Therefore, \(D < 0\).

For \(f\), \(D > 0\). For \(g\), \(D < 0\).

- Identify the parameter as the leading coefficient before forming the discriminant. - Translate “exactly one real solution” into a condition on \(D\). - Check that the resulting value of \(k\) is nonzero as required.

1. Here \(D(k)=12^2-4\cdot k\cdot18=144-72k\). 2. Exactly one real solution requires \(D(k)=0\). 3. Solve \(144-72k=0\): \(k=2\).

\(D(k)=144-72k\); \(144-72k=0\), so \(k=2\).

- Both the linear coefficient and constant term depend on the same parameter. - A tangent x-axis contact gives a zero discriminant. - After forming the parameter equation, factor it rather than solving the original quadratic in \(x\).

1. The discriminant is \(D(a)=a^2-4a\). 2. Touching once requires \(D(a)=0\), so \(a^2-4a=0\). 3. Factor: \(a(a-4)=0\), giving \(a=0\) or \(a=4\).

\(D(a)=a^2-4a\); \(a(a-4)=0\), so \(a=0\) or \(a=4\).

- Calculate \(b^2 - 4ac\) in terms of \(q\). - Match positive, zero, and negative discriminants with the possible numbers of real solutions. - Solve each resulting inequality or equation for \(q\).

a) The discriminant is \(D=6^2-4\cdot1\cdot q=36-4q=4(9-q)\). For exactly two real solutions, require \(D>0\): \(4(9-q)>0\), so \(q<9\). b) For exactly one real solution, require \(D=0\): \(4(9-q)=0\), so \(q=9\). c) For no real solutions, require \(D<0\): \(4(9-q)<0\), so \(q>9\).

a) \(q < 9\) b) \(q = 9\) c) \(q > 9\)

- Put each equation in standard form before reading \(a\), \(b\), and \(c\). - For part b), expand the squared binomial only for the discriminant calculation. - After classifying the roots, use an efficient exact method to find them.

1. a) \(D=-100<0\), so there are no real solutions. 2. b) Expanding gives \(x^2-2\sqrt2x+2=0\), whose discriminant is \(0\); the repeated solution is \(x=\sqrt2\). 3. c) \(D=144>0\), so there are two real solutions. Solving gives \(x=-2\) and \(x=2\).

a) \(D=-100\); no real solutions b) \(D=0\); one real solution, \(x=\sqrt2\) c) \(D=144\); two real solutions, \(x=-2,2\)

- Treat \(a\) as the parameter after moving it to the left side. - Use the sign of \(4a\) rather than taking square roots first. - Repeat the same sign analysis for \(-4c\).

1. Rewrite as \(x^2-a=0\). Its discriminant is \(D=4a\). 2. If \(a<0\), then \(D<0\) and there are no real solutions; for example, \(x^2=-4\). 3. If \(a=0\), then \(D=0\) and there is one real solution, \(x=0\). 4. If \(a>0\), then \(D>0\) and there are two real solutions; for example, \(x^2=9\) gives \(x=\pm3\). 5. For \(x^2+c=0\), \(D=-4c\). No real zeros require \(-4c<0\), so \(c>0\).

For \(x^2=a\), \(D=4a\). The claim is false: \(a<0\) gives no real solutions, \(a=0\) gives one, and \(a>0\) gives two. Examples are \(x^2=-4\), \(x^2=0\), and \(x^2=9\), respectively. For \(f(x)=x^2+c\), \(D=-4c\); no real zeros require \(D<0\), so \(c>0\).

- Translate the vertex condition into a repeated-root condition first. - Form \(D\) in terms of \(k\) before trying to rewrite the trinomial as a square. - After finding \(k\), use each perfect-square form to read the vertex.

1. A vertex on the x-axis gives one repeated real zero, so \(D=0\). 2. \(D=k^2-36\), so \(k^2-36=0\), giving \(k=6\) or \(k=-6\). 3. For \(k=6\), \(f_k=(x+3)^2\), vertex \((-3,0)\). 4. For \(k=-6\), \(f_k=(x-3)^2\), vertex \((3,0)\).

\(D=k^2-36=0\), so \(k=6\) or \(k=-6\). \(k=6\): vertex \((-3,0)\) \(k=-6\): vertex \((3,0)\)

- Keep \(c\) symbolic in the discriminant for part a). - Exactly one real solution corresponds to \(D=0\). - Reuse the same discriminant expression in part b) rather than starting over.

1. \(D(c)=64-4c\). 2. a) Set \(64-4c=0\), so \(c=16\). The repeated solution is \(x=4\). 3. b) \(D=64-80=-16<0\), so there are no real solutions.

a) \(D(c)=64-4c\); \(64-4c=0\) gives \(c=16\), with repeated solution \(x=4\). b) \(D=-16\); no real solutions.

- What must the vertex's \(x\)-coordinate be to lie on the \(y\)-axis? - How many real zeros does a quadratic have when its graph only touches the \(x\)-axis? - Which expression determines the number of real solutions of a quadratic equation?

a) The vertex x-coordinate is \(-\frac{b}{2a}=-\frac{k}{4}\). For the vertex to lie on the y-axis, set \(-\frac{k}{4}=0\), giving \(k=0\). b) Tangency to the x-axis requires discriminant \(0\): \(k^2-4\cdot2\cdot8=0\). Thus, \(k^2-64=0\), so \(k=8\) or \(k=-8\). In either case, the single x-axis contact point is the vertex, so the vertex lies on the x-axis.

a) \(k = 0\) b) \(k = 8\) or \(k = -8\); in either case, the vertex lies on the \(x\)-axis.

- Set the height function equal to each target height. - Move all terms to one side before calculating the discriminant. - Use the sign of the discriminant to determine the number of times.

a) Set \(h(t)=21\): \(-16t^2+32t-17=0\). The discriminant is \(D=32^2-4\cdot(-16)\cdot(-17)=1024-1088=-64\). Since \(D<0\), the ball does not reach \(21\,\text{ft}\). b) Set \(h(t)=20\): \(-16t^2+32t-16=0\). The discriminant is \(D=32^2-4\cdot(-16)\cdot(-16)=1024-1024=0\). Therefore, the ball is exactly \(20\,\text{ft}\) high at one time.

a) No; the discriminant is \(-64\). b) The ball is at \(20\,\text{ft}\) at exactly one time.

- Set the two function expressions equal. - Tangency corresponds to one repeated solution. - Use the discriminant to find \(n\), then substitute to find the point.

1. Set the functions equal: \(-x^2 + 3 = 2x + n\). 2. Rearrange: \(x^2 + 2x + (n - 3) = 0\). 3. Tangency means exactly one intersection, so the discriminant must be \(0\). 4. Calculate \(D = 2^2 - 4 \cdot 1 \cdot (n - 3) = 16 - 4n\). 5. Set \(16 - 4n = 0\), giving \(n = 4\). 6. Then \(x^2 + 2x + 1 = (x + 1)^2 = 0\), so \(x = -1\). The y-coordinate is \(g(-1) = 2\). 7. The point of tangency is \((-1, 2)\).

\(n = 4\); point of tangency \((-1, 2)\)

- Substitute the given value of \(c\) before solving parts a) and b). - For part c), write the discriminant in terms of \(c\). - Relate the sign of the discriminant to the number of real solutions.

a) When \(c=50\), \(2x^2-50=0\), so \(x^2=25\). Thus, \(x=-5\) or \(x=5\). b) When \(c=10\), \(2x^2-10=0\), so \(x^2=5\). Thus, \(x=-\sqrt{5}\) or \(x=\sqrt{5}\). c) For \(2x^2-c=0\), the discriminant is \(D=0^2-4\cdot2\cdot(-c)=8c\). If \(c>0\), then \(D>0\) and there are two real solutions; if \(c=0\), then \(D=0\) and there is one real solution; if \(c<0\), then \(D<0\) and there are no real solutions.

a) \(\{-5, 5\}\) b) \(x = -\sqrt{5}\) or \(x = \sqrt{5}\) c) Two real solutions when \(c > 0\), one real solution when \(c = 0\), and no real solutions when \(c < 0\).

- Write the discriminant as an expression in \(c\). - Use \(D = 0\) for exactly one real solution. - Use \(D < 0\) to choose a value for part c).

a) The discriminant is \(D=8^2-4\cdot2\cdot c=64-8c\). When \(c=5\), \(D=64-8\cdot5=24>0\), so the equation has two real solutions. b) For exactly one real solution, set \(D=0\): \(64-8c=0\), so \(c=8\). c) No real solutions require \(D<0\): \(64-8c<0\), so \(c>8\). One possible value is \(c=10\).

a) \(D = 24\); two real solutions b) \(c = 8\) c) One possible value is \(c = 10\).

- Check whether equation (I) is a perfect-square trinomial. - For equation (II), identify \(a\), \(b\), and \(c\). - Recall how the sign of the discriminant determines the number of real solutions.

a) \(x^2-6x+9=(x-3)^2\). Therefore, \((x-3)^2=0\), so \(x=3\). b) With \(a=1\), \(b=-6\), and \(c=8\), the quadratic formula gives \(x=\frac{6\pm\sqrt{(-6)^2-4\cdot1\cdot8}}{2}=\frac{6\pm2}{2}\), so \(x=4\) or \(x=2\). c) Equation (I) has discriminant \((-6)^2-4\cdot1\cdot9=0\), so it has one repeated real solution. Equation (II) has discriminant \((-6)^2-4\cdot1\cdot8=4>0\), so it has two distinct real solutions.

a) \(x = 3\) b) \(x \in \{2, 4\}\) c) Equation (I) has discriminant \(0\), while equation (II) has a positive discriminant.

- Set the function expressions equal. - Look for a perfect-square trinomial. - A repeated solution corresponds to one common point.

1. Set the functions equal: \(x^2 - 4x + 3 = 2x - 6\). 2. Rearrange: \(x^2 - 6x + 9 = 0\). 3. Factor: \((x - 3)^2 = 0\), so there is one repeated solution, \(x = 3\). 4. The repeated solution means the graphs have exactly one common point, so the line is tangent to the parabola. 5. Calculate the y-coordinate: \(g(3) = 2 \cdot 3 - 6 = 0\). The point of tangency is \((3, 0)\).

The point of tangency is \((3, 0)\). The equation for the intersection has the repeated solution \(x = 3\).

- Set the function expressions equal. - Expand the squared binomial and write a quadratic equation. - Use the discriminant to determine whether real intersections exist.

1. Set the functions equal: \((x - 2)^2 + 3 = x\). 2. Expand and rearrange: \(x^2 - 4x + 4 + 3 = x\), so \(x^2 - 5x + 7 = 0\). 3. The discriminant is \(D = (-5)^2 - 4 \cdot 1 \cdot 7 = -3\). 4. Since \(D < 0\), there are no real solutions, so the graphs have no common points.

The graphs have no common points.

- Put each function into standard form before computing its discriminant. - Use \(D\) only for the number of real zeros; then choose an efficient exact method for their values. - A zero discriminant corresponds to a repeated root.

1. a) \(D=100-84=16>0\), so there are two real zeros. Factoring gives \(x=-7,-3\). 2. b) Rewrite as \(-3x^2-6x-9\). Then \(D=36-108=-72<0\), so there are no real zeros. 3. c) \(D=16-16=0\), so there is one distinct real zero. The equation is equivalent to \((x-4)^2=0\), so the zero is \(x=4\) with multiplicity \(2\).

a) \(D=16\); two real zeros: \(x=-7,-3\) b) \(D=-72\); no real zeros c) \(D=0\); one distinct real zero: \(x=4\), multiplicity \(2\)

- Substitute \(k = 15\) before solving part a). - Rewrite the general equation in standard form. - Use the sign of its discriminant for parts b) and c).

a) When \(k=15\), \(3x^2-12=15\), so \(3x^2=27\) and \(x^2=9\). Thus, \(x=-3\) or \(x=3\). b) Rewrite the general equation as \(3x^2-(k+12)=0\). Its discriminant is \(D=12(k+12)\). Exactly one real solution requires \(D=0\), so \(k=-12\), giving \(x=0\). c) No real solutions require \(D<0\), so \(k+12<0\) and \(k<-12\).

a) \(\{-3, 3\}\) b) \(k = -12\) c) \(k < -12\)

- Substitute the given solution into the equation in part a). - Treat \(c = 0\) separately before dividing by \(c\). - For \(c \ne 0\), determine when \(\frac{18}{c}\) is negative.

a) Substitute \(x=3\): \(c(3)^2-18=0\), so \(9c=18\) and \(c=2\). Then \(2x^2-18=0\), so \(x^2=9\); the second solution is \(x=-3\). b) When \(c=0\), the equation becomes \(-18=0\), which is false for every \(x\). c) For \(c\ne0\), \(x^2=\frac{18}{c}\), which has no real solution when \(c<0\). Including the separate case \(c=0\), there is no real solution for \(c\le0\).

a) \(c = 2\); the second solution is \(x = -3\). b) The equation becomes the false statement \(-18 = 0\). c) \(c \le 0\)

- Clear both denominators before trying to identify \(a\), \(b\), and \(c\). - Keep careful track of the subtraction in the second equation. - The requested conclusion comes from the sign of \(D\), not from solving for the roots.

1. For (1), multiply by \(15\): \(3(x^2+x+1)+5(2x^2-2x-2)=45\), so \(13x^2-7x-52=0\). 2. Its discriminant is \(D=(-7)^2-4\cdot13\cdot(-52)=2753>0\), so there are two distinct real solutions. 3. For (2), multiply by \(15\): \(3(x^2+x+1)-5(2x^2-2x-2)=45\), so \(7x^2-13x+32=0\). 4. Its discriminant is \(D=(-13)^2-4\cdot7\cdot32=-727<0\), so there are no real solutions.

(1) \(13x^2-7x-52=0\); \(D=2753>0\); two distinct real solutions. (2) \(7x^2-13x+32=0\); \(D=-727<0\); no real solutions.

- Use the standard discriminant \(b^2 - 4ac\). - Set the discriminant equal to \(0\) for one real zero. - Use the opening direction to interpret a negative discriminant graphically.

a) The discriminant is \(D=(-6)^2-4\cdot1\cdot k=36-4k\). b) For exactly one real zero, set \(D=0\): \(36-4k=0\), so \(k=9\). Then \(f_k(x)=x^2-6x+9=(x-3)^2\), so the vertex is \((3,0)\). c) When \(k=10\), \(D=36-4\cdot10=-4\), so there are no real zeros. Because the parabola opens upward, its graph lies entirely above the x-axis.

a) \(D = 36 - 4k\) b) \(k = 9\); vertex \((3, 0)\) c) No real zeros; the graph lies entirely above the x-axis.

- Set the two expressions for \(y\) equal. - Use the discriminant to test for intersections. - Write the parabola in vertex form to interpret the result geometrically.

1. Set the two expressions for \(y\) equal: \(x^2 - 4x + 6 = 1\). 2. Rearrange: \(x^2 - 4x + 5 = 0\). 3. The discriminant is \(D = (-4)^2 - 4 \cdot 1 \cdot 5 = -4\). Since \(D < 0\), there are no real intersection points. 4. Complete the square: \(y = x^2 - 4x + 6 = (x - 2)^2 + 2\). The vertex is \((2, 2)\). 5. The parabola opens upward and has minimum value \(2\), so it lies entirely above the line \(y = 1\).

The system has no solution. The parabola has vertex \((2, 2)\) and lies entirely above \(y = 1\).

- Set the two function expressions equal. - Clear decimals before calculating the discriminant. - A negative discriminant means the line and parabola do not intersect.

1. Set the functions equal: \(-0.5x^2 + 2x - 4 = 0.5x - 1\). 2. Rearrange: \(-0.5x^2 + 1.5x - 3 = 0\). 3. Multiply by \(-2\): \(x^2 - 3x + 6 = 0\). 4. The discriminant is \(D = (-3)^2 - 4 \cdot 1 \cdot 6 = -15\). 5. Since \(D < 0\), the graphs have no common points. The line does not intersect the parabola.

The graphs have no common points; the line does not intersect the parabola.

- Set the function expressions equal. - For part b), express the discriminant in terms of \(m\). - A tangent gives one repeated intersection x-value.

1. When \(m = -2\), set \(x^2 - 4x + 3 = -2x + 3\). 2. This simplifies to \(x^2 - 2x = x(x - 2) = 0\), so \(x = 0\) or \(x = 2\). 3. Using \(y = -2x + 3\), the intersection points are \((0, 3)\) and \((2, -1)\). 4. In general, setting the functions equal gives \(x^2 - (4 + m)x = 0\). 5. Its discriminant is \(D = (4 + m)^2\). Tangency requires \(D = 0\), so \(m = -4\). 6. Then the repeated solution is \(x = 0\), and \(y = 3\). The point of tangency is \((0, 3)\).

a) \((0, 3)\) and \((2, -1)\) b) \(m = -4\); point of tangency \((0, 3)\)

- Set the line equal to the parabola. - Require the resulting quadratic to have discriminant \(0\). - Use the repeated x-value to find the y-coordinate.

1. Set the functions equal: \(x^2 - 4x + 5 = 2x + c\). 2. Rearrange: \(x^2 - 6x + (5 - c) = 0\). 3. The discriminant is \(D = (-6)^2 - 4 \cdot 1 \cdot (5 - c) = 16 + 4c\). 4. Tangency requires \(D = 0\), so \(16 + 4c = 0\) and \(c = -4\). 5. Then \(x^2 - 6x + 9 = (x - 3)^2 = 0\), so \(x = 3\). 6. The y-coordinate is \(2 \cdot 3 - 4 = 2\). The point of tangency is \((3, 2)\).

\(c = -4\); point of tangency \((3, 2)\)

- Put the equation in standard form. - Express its discriminant in terms of \(d\). - Match each sign of the discriminant with the number of real solutions.

a) Write the equation as \(5x^2+d-20=0\). Its discriminant is \(D=0^2-4\cdot5(d-20)=20(20-d)\). Exactly two real solutions require \(D>0\), so \(d<20\). b) Exactly one real solution requires \(D=0\), so \(d=20\). Then \(x=0\). c) No real solutions require \(D<0\), so \(d>20\).

a) \(d < 20\) b) \(d = 20\) c) \(d > 20\)

- Write the equation in standard quadratic form. - Calculate the discriminant in terms of \(a\) and \(b\). - Use the facts that \(a > 0\) and a real solution requires a nonnegative discriminant.

1. Expand and write the equation in standard form: \(ax^2 + a - b = 0\). 2. Since \(a > 0\), this is a quadratic equation with discriminant \(D = 0^2 - 4a(a - b) = 4a(b - a)\). 3. A real solution exists when \(D \ge 0\). Because \(a > 0\), this requires \(b - a \ge 0\), or \(b \ge a\). 4. Therefore, the claim is false. There is no real solution when \(b < a\). 5. For example, let \(a = 2\) and \(b = 1\). Then \(2(x^2 + 1) = 1\), which gives \(x^2 = -\frac{1}{2}\), so there is no real solution.

The claim is false. There is no real solution when \(b < a\). One counterexample is \(2(x^2 + 1) = 1\).

- Express the discriminant in terms of \(c\). - Set it equal to \(0\) in part a). - Determine the sign of the discriminant when \(c\) is greater than the value you found in part a).

a) The discriminant is \(D=(-10)^2-4\cdot1\cdot c=100-4c\). Exactly one real solution requires \(D=0\), so \(100-4c=0\) and \(c=25\). b) If \(c>25\), then \(4c>100\), so \(D=100-4c<0\). Therefore, the equation has no real solutions.

a) \(c = 25\) b) For \(c > 25\), the equation has no real solutions.

- Connect the square of a binomial to one repeated real zero. - Keep the entire linear coefficient \(m+3\) together when squaring it. - After finding the parameter values, verify them by expanding the claimed squares.

1. A perfect-square quadratic has a repeated root, so \(D=0\). 2. \(D(m)=(m+3)^2-400\). 3. Set \((m+3)^2-400=0\), giving \(m=17\) or \(m=-23\). 4. The corresponding expressions are \((2x+5)^2\) and \((2x-5)^2\), respectively.

\(D(m)=(m+3)^2-400=0\) \(m=17\): \((2x+5)^2\) \(m=-23\): \((2x-5)^2\)

- Expand the product in \(4ac\) carefully before simplifying the discriminant. - A zero discriminant is necessary for a nonconstant quadratic to be one squared linear factor up to scale. - Check that the resulting leading coefficient is not zero before writing the square form.

1. The discriminant is \(D(k)=(2k)^2-4(k-1)(k+3)=-8k+12\). 2. Set \(-8k+12=0\), so \(k=\frac32\). 3. The leading coefficient is then \(\frac12\ne0\), and \(f(x)=\frac12x^2+3x+\frac92=\frac12(x+3)^2\).

\(D(k)=-8k+12\); \(D=0\) gives \(k=\frac32\). Verification: \(f(x)=\frac12(x+3)^2\).

- When does a quadratic equation have exactly one real solution? - How is a vertex on the \(x\)-axis related to the number of zeros? - What does the sign of the leading coefficient tell you about a parabola? - How does \((x - h)^2 + k\) expand into standard form?

1) The graph touches the x-axis at exactly one point when the quadratic equation has one real solution, so its discriminant must be zero: \(p^2-4q=0\). Equivalently, \(q=\frac{p^2}{4}\). 2) A vertex at \((5,0)\) gives \(f(x)=(x-5)^2=x^2-10x+25\). Thus, \(p=-10\) and \(q=25\). 3) The leading coefficient is \(1>0\), so the parabola opens upward. Its vertex is a minimum, and there is no maximum because the function values increase without bound.

1) \(p^2 - 4q = 0\), equivalently \(q = \frac{p^2}{4}\) 2) \(p = -10\) and \(q = 25\) 3) The positive leading coefficient makes the parabola open upward, so its vertex is a minimum and there is no maximum.

- Set the function equal to \(0\) and compute the discriminant. - The sign of the discriminant determines the number of x-intercepts. - Solve each condition for \(k\).

1. Set \(g_k(x)=0\): \(-x^2+4x+k=0\). The discriminant is \(D=4^2-4\cdot(-1)\cdot k=16+4k\). 2. There are two x-intercepts when \(D>0\): \(16+4k>0\), so \(k>-4\). 3. There is one x-intercept when \(D=0\): \(16+4k=0\), so \(k=-4\). 4. There are no x-intercepts when \(D<0\): \(16+4k<0\), so \(k<-4\).

Two x-intercepts for \(k>-4\) One x-intercept for \(k=-4\) No x-intercepts for \(k<-4\)

- Count each graph’s x-intercepts. - Connect two, one, or zero x-intercepts with the sign of the discriminant. - For part b), identify \(a\), \(b\), and \(c\) from the equation.

a) The graph of \(f\) crosses the x-axis twice, so \(f(x)=0\) has two real solutions and \(D>0\). The graph of \(g\) stays above the x-axis, so \(g(x)=0\) has no real solutions and \(D<0\). The graph of \(h\) touches the x-axis once, so \(h(x)=0\) has one real solution and \(D=0\). b) For \(f(x)=-x^2+4x-2\), \(D=4^2-4\cdot(-1)\cdot(-2)=16-8=8\). Since \(8>0\), the calculation agrees with the two x-intercepts shown in the graph.

a) \(f\): two real solutions and \(D > 0\) \(g\): no real solutions and \(D < 0\) \(h\): one real solution and \(D = 0\) b) \(D = 8\), which agrees with the graph.

- Find the line’s slope from two points. - Set the line equal to the parabola. - A tangent produces a repeated solution.

1. The parabola is \(f(x) = x^2\). 2. The line passes through \((1, 0)\) and \((2, 4)\), so its slope is \(4\) and its equation is \(g(x) = 4x - 4\). 3. Set the functions equal: \(x^2 = 4x - 4\). 4. Move all terms to one side: \(x^2 - 4x + 4 = 0\). 5. This is \((x - 2)^2 = 0\), so it has the single repeated solution \(x = 2\).

\(x^2 - 4x + 4 = 0\)

- Use symmetry to identify the vertex. - Determine the opening direction from the three points. - Decide where the vertex must be for two x-intercepts.

a) Points \(A\) and \(C\) are symmetric about the y-axis and have the same y-coordinate, so \(B(0, 1)\) is the vertex. The parabola opens upward and its vertex lies above the x-axis. Therefore, it has no real zeros and \(D < 0\). b) To make \(D > 0\), the vertex must lie below the x-axis. The y-coordinate of \(B\) must be decreased from \(1\) to a value less than \(0\), so it must be decreased by more than \(1\).

a) \(D < 0\) b) The y-coordinate must be decreased by more than \(1\).

- Convert graph intersections into equal function outputs first. - Form the discriminant only after the intersection equation is in standard form. - “Touch or intersect” means at least one real solution, not necessarily two.

1. Set the outputs equal: \(2x^2+4x+(7-t)=0\). 2. Its discriminant is \(D=4^2-4\cdot2\cdot(7-t)=8t-40\). 3. The graphs have at least one common point exactly when \(D\ge0\). 4. Thus, \(8t-40\ge0\), so \(t\ge5\).

Intersection equation: \(2x^2+4x+(7-t)=0\) Discriminant: \(D=8t-40\) Condition: \(8t-40\ge0\), so \(t\ge5\).

- Set the function expressions equal. - Use the discriminant to test for common points. - For part b), require the discriminant to be nonnegative.

1. Set \(f(x) = g(x)\): \(x^2 + 1 = x - 1\), so \(x^2 - x + 2 = 0\). 2. Its discriminant is \(D = (-1)^2 - 4 \cdot 1 \cdot 2 = -7\). Since \(D < 0\), the graphs do not intersect. 3. For the shifted line, set \(x^2 + 1 = x - 1 + d\), giving \(x^2 - x + (2 - d) = 0\). 4. Touching or intersecting requires \(D \ge 0\): \(1 - 4(2 - d) = 4d - 7 \ge 0\). 5. Therefore, \(d \ge \frac{7}{4}\), so the minimum upward shift is \(\frac{7}{4}\) units.

a) The discriminant is \(-7\), so there are no intersections. b) The minimum shift is \(d = \frac{7}{4}\).

- Keep the sign of \(b=-1\) attached when substituting into the formula. - Multiply the conjugate exact roots before converting anything to decimals. - Exactly one real root corresponds to a zero discriminant.

1. With \(a=1\), \(b=-1\), and \(c=-5\), the quadratic formula gives \(x=\frac{1\pm\sqrt{21}}{2}\). 2. Their product is \(\frac{(1+\sqrt{21})(1-\sqrt{21})}{4}=\frac{1-21}{4}=-5\). 3. For \(x^2-x+c=0\), exactly one real solution requires \(D=(-1)^2-4c=0\). 4. Thus, \(c=\frac14\).

a) \(x=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-5)}}{2\cdot1}=\frac{1\pm\sqrt{21}}{2}\) b) \(\frac{1+\sqrt{21}}{2}\cdot\frac{1-\sqrt{21}}{2}=-5\) c) \(D=1-4c=0\), so \(c=\frac14\).

- Express the discriminant in terms of \(p\). - Match each required number of real solutions with a sign condition on the discriminant. - Solve the resulting quadratic inequalities carefully.

a) The discriminant is \(D=p^2-4\cdot1\cdot9=p^2-36\). For exactly one real solution, require \(D=0\): \(p^2-36=0\), so \(p=-6\) or \(p=6\). b) For no real solutions, require \(D<0\): \(p^2<36\), so \(-6<p<6\). c) For two distinct real solutions, require \(D>0\): \(p^2>36\), so \(p<-6\) or \(p>6\).

a) \(p = -6\) or \(p = 6\) b) \(-6 < p < 6\) c) \(p < -6\) or \(p > 6\)

- Multiply by the least common denominator to clear the fractions. - Combine the \(x^2\)-terms carefully. - For part b, use the condition that a quadratic has exactly one real solution when its discriminant is zero.

a) Multiply by \(12\): \(3(3x^2+5)-4(2x^2-1)=24\). Simplifying gives \(x^2=5\), so \(x=\pm\sqrt{5}\). b) Multiply the parameter equation by \(12\): \(x^2+19=12c\), so \(x^2+19-12c=0\). Exactly one real solution requires discriminant \(0\): \(-4(19-12c)=0\), so \(c=\frac{19}{12}\).

a) \(x \in \{-\sqrt{5}, \sqrt{5}\}\) b) \(c = \frac{19}{12}\)

- Solve the linear equation for one variable. - Substitute to obtain a quadratic equation in one variable. - Require its discriminant to equal \(0\).

1. Solve the second equation for \(y\): \(y = 10 - x\). 2. Substitute into \(xy = c\): \(x(10 - x) = c\). 3. Rearrange: \(x^2 - 10x + c = 0\). 4. The discriminant is \(D = (-10)^2 - 4 \cdot 1 \cdot c = 100 - 4c\). 5. For exactly one value of \(x\), set \(D = 0\): \(100 - 4c = 0\), so \(c = 25\). 6. Then \((x - 5)^2 = 0\), so \(x = 5\), and \(y = 10 - 5 = 5\).

When \(c = 25\), the system has the single solution \((5, 5)\).

- Isolate one variable in the linear equation. - Substitute into the product equation and factor the resulting quadratic. - Compare the two ordered pairs when \(s = 0\) and when \(s \ne 0\).

1. Solve the first equation for \(y\): \(y = 7s - x\). 2. Substitute into the second equation: \(x(7s - x) = 10s^2\), so \(x^2 - 7sx + 10s^2 = 0\). 3. Factor: \((x - 2s)(x - 5s) = 0\). Thus, \(x = 2s\) or \(x = 5s\). 4. The corresponding values of \(y\) are \(5s\) and \(2s\), giving \((2s, 5s)\) and \((5s, 2s)\). 5. When \(s = 0\), both ordered pairs become \((0, 0)\). When \(s \ne 0\), \(2s \ne 5s\), so the ordered pairs are distinct.

a) \(\{(2s, 5s), (5s, 2s)\}\) b) For \(s = 0\), both expressions give \((0, 0)\). For \(s \ne 0\), they give two distinct ordered pairs.

- Set the two expressions for \(y\) equal. - Require the resulting quadratic to have a zero discriminant. - Substitute the repeated x-value to find \(y\).

1. Set the expressions for \(y\) equal: \(x^2 - 4x + k = 2x - 5\). 2. Rearrange: \(x^2 - 6x + (k + 5) = 0\). 3. The discriminant is \(D = (-6)^2 - 4 \cdot 1 \cdot (k + 5) = 16 - 4k\). 4. For exactly one solution, set \(D = 0\): \(16 - 4k = 0\), so \(k = 4\). 5. Then \(x^2 - 6x + 9 = (x - 3)^2 = 0\), so \(x = 3\). 6. Substitute into the line: \(y = 2 \cdot 3 - 5 = 1\).

a) \(k = 4\) b) \((3, 1)\)

- Think of \(x\) and \(y\) as two roots with a known sum and product. - Build the monic quadratic from those two symmetric quantities. - Exactly one ordered pair occurs when the two roots merge into one repeated value.

1. Numbers with sum \(12\) and product \(k\) are roots of \(z^2-12z+k=0\). 2. For \(k=32\), \(z^2-12z+32=(z-8)(z-4)=0\), so the ordered pairs are \((8,4)\) and \((4,8)\). 3. The discriminant is \(D(k)=144-4k\). 4. Exactly one ordered pair occurs when the two coordinate values coincide, so \(D(k)=0\). This gives \(k=36\), with pair \((6,6)\).

a) \(\{(8,4),(4,8)\}\) b) Quadratic: \(z^2-12z+k=0\); \(D(k)=144-4k\); \(D=0\) gives \(k=36\), with \((6,6)\).

- Write the discriminant as a quadratic expression in \(k\). - Factor that expression. - Use a sign chart or the zeros of the factored expression to solve each inequality.

1. The discriminant is \(D = (k - 2)^2 - 4 \cdot 1 \cdot (k + 1)\). 2. Simplify: \(D = k^2 - 4k + 4 - 4k - 4 = k^2 - 8k = k(k - 8)\). 3. Exactly one real solution occurs when \(D = 0\), so \(k = 0\) or \(k = 8\). 4. No real solutions occur when \(D < 0\). The product \(k(k - 8)\) is negative between its zeros, so \(0 < k < 8\). 5. Two distinct real solutions occur when \(D > 0\), so \(k < 0\) or \(k > 8\).

No real solutions: \(0 < k < 8\) Exactly one real solution: \(k = 0\) or \(k = 8\) Two distinct real solutions: \(k < 0\) or \(k > 8\)

- Use the positive leading coefficient and the absence of real zeros. - Write and factor the discriminant as an expression in \(k\). - Determine where the factored expression is negative.

1. The leading coefficient is positive, so the expression is positive for every real \(x\) exactly when the corresponding parabola has no real zeros. 2. Its discriminant is \(D = k^2 - 4 \cdot 1 \cdot (k + 3) = k^2 - 4k - 12\). 3. Require \(D < 0\): \(k^2 - 4k - 12 < 0\). 4. Factor: \((k - 6)(k + 2) < 0\). 5. This product is negative between its zeros, so \(-2 < k < 6\).

\(-2 < k < 6\)

- First identify when the coefficient of \(x^2\) is \(0\). - Treat that linear case separately. - For all other values, use the discriminant.

1. If \(a = 2\), the quadratic term disappears and the equation becomes \(4x + 1 = 0\). It has one solution, \(x = -\frac{1}{4}\). 2. If \(a \ne 2\), the equation is quadratic. Its discriminant is \(D = 4^2 - 4 \cdot (a - 2) \cdot 1 = 24 - 4a\). 3. When \(D = 0\), \(24 - 4a = 0\), so \(a = 6\). The quadratic has one repeated real solution. 4. When \(D > 0\), \(a < 6\). Excluding the linear case \(a = 2\), the equation has two distinct real solutions for \(a < 6\) and \(a \ne 2\). 5. When \(D < 0\), \(a > 6\), so there are no real solutions.

One real solution: \(a = 2\) or \(a = 6\) Two distinct real solutions: \(a < 6\) and \(a \ne 2\) No real solutions: \(a > 6\)

- Set the line and parabola equal. - Use the discriminant to count intersections. - Tangency means a zero discriminant.

1. Set the expressions equal: \(x^2 + 2x + 2 = -2x - 3\), so \(x^2 + 4x + 5 = 0\). 2. The discriminant is \(D = 4^2 - 4 \cdot 1 \cdot 5 = -4\), so the original system has no real solutions. 3. For the shifted line, setting the functions equal gives \(x^2 + 4x + (2 - c) = 0\). 4. Its discriminant is \(D = 4^2 - 4 \cdot 1 \cdot (2 - c) = 8 + 4c\). 5. Tangency requires \(D = 0\), so \(8 + 4c = 0\) and \(c = -2\). 6. If \(c > -2\), then \(D > 0\), so the line and parabola have two intersection points.

1) No real solutions 2) \(c = -2\) 3) Two intersections when \(c > -2\)

- Use the vertex and one point to determine \(f\). - Set \(f(x)\) equal to \(g_k(x)\). - Treat a possible linear case separately, then require a negative discriminant.

1. From the graph, \(f\) has vertex \((2, 2)\) and passes through \((0, 0)\). 2. Write \(f(x) = a(x - 2)^2 + 2\). Substituting \((0, 0)\) gives \(0 = 4a + 2\), so \(a = -\frac{1}{2}\). Thus, \(f(x) = -\frac{1}{2}(x - 2)^2 + 2 = -\frac{1}{2}x^2 + 2x\). 3. Set the functions equal: \(-\frac{1}{2}x^2 + 2x = kx^2 - 4\). 4. Rearrange: \(\left(k + \frac{1}{2}\right)x^2 - 2x - 4 = 0\). 5. If \(k = -\frac{1}{2}\), the equation is linear and has a real solution. Otherwise, its discriminant is \(D = (-2)^2 - 4 \cdot \left(k + \frac{1}{2}\right) \cdot (-4) = 16k + 12\). 6. No common points require \(D < 0\): \(16k + 12 < 0\), so \(k < -\frac{3}{4}\).

\(k < -\frac{3}{4}\)

- Determine the equation of the parabola from its vertex and another point. - Notice the common point through which every line in the family passes. - A substitution for \(x-2\) can simplify the intersection equation. - Use the discriminant condition for a quadratic equation with no real solutions.

1. The parabola has vertex \((2,1)\) and passes through \((0,2)\). Write \(f(x)=a(x-2)^2+1\). Substituting \((0,2)\) gives \(2=4a+1\), so \(a=0.25\). Thus, \(f(x)=0.25(x-2)^2+1\). 2. Set the functions equal: \(0.25(x-2)^2+1=m(x-2)\). 3. Let \(u=x-2\). The equation becomes \(0.25u^2-mu+1=0\). 4. Its discriminant is \(D=(-m)^2-4 \cdot 0.25 \cdot 1=m^2-1\). 5. There are no intersections when \(D<0\). Therefore, \(m^2-1<0\), which gives \(-1<m<1\).

\(-1<m<1\)

- Read each vertex directly from vertex form. - Eliminate the parameter to find the vertex locus. - For a fixed point \((x, y)\), treat the equation as a quadratic equation in \(a\). - Use the discriminant to decide when a real parameter value exists.

a) The equation is in vertex form, so the vertex is \((a,2a)\). Eliminating \(a\) gives the vertex locus \(y=2x\). b) A fixed point \((x,y)\) lies on a graph in the family exactly when \(y=(x-a)^2+2a\) has a real solution for \(a\). Rearranging gives \(a^2+(2-2x)a+x^2-y=0\). Its discriminant is \(\Delta=(2-2x)^2-4(x^2-y)=4(1-2x+y)\). A real \(a\) exists exactly when \(\Delta\ge0\), or \(y\ge2x-1\). Therefore, points with \(y<2x-1\) lie on no graph in the family.

a) \(y=2x\) b) All points \((x, y)\) such that \(y<2x-1\).

- Read a vertex and one additional clear point from \(p\) to identify its equation. - Use the required vertex of \(h\) to write a one-parameter family. - Intersections become solutions of one equation obtained by setting the two functions equal. - Check separately whether the quadratic intersection equation can degenerate to a linear equation.

1. From the graph, \(p(x)=0.5(x-2)^2+1=0.5x^2-2x+3\). 2. A quadratic with vertex \((0,5)\) has form \(h(x)=ax^2+5\), with \(a\ne0\). 3. Intersections satisfy \(ax^2+5=0.5x^2-2x+3\), so \((a-0.5)x^2+2x+2=0\). 4. If \(a=0.5\), this becomes the linear equation \(2x+2=0\), so the graphs intersect and that value is excluded. 5. For \(a\ne0.5\), no common point requires discriminant \(D=2^2-4(a-0.5)(2)<0\). 6. Thus, \(4-8a+4<0\), so \(8-8a<0\), giving \(a>1\).

\(h(x)=ax^2+5\) for every \(a>1\).