Free math worksheets

- A turning point is where the graph changes from increasing to decreasing or from decreasing to increasing. - Trace the graph from left to right and count each direction change.

1. Reading from left to right, the graph changes direction three times: near \(x=-2\), \(x=0\), and \(x=2\). 2. At \(x=0\), the graph changes from increasing to decreasing, so that turning point is a local maximum.

The graph has \(3\) turning points. The turning point at \(x=0\) is a local maximum.

- Focus on how the direction changes at \(x=5\). - Compare “increasing then decreasing” with the definition of a local maximum or minimum.

1. The graph rises as it approaches \(x=5\). 2. It falls after \(x=5\), so the direction change is from increasing to decreasing. Therefore, the turning point is a local maximum.

A local maximum.

- Use the vertex of the quadratic to find its minimum. - Determine what taking reciprocals does to the smallest and largest positive values. - Compare each function at \(-x\) and \(x\).

1. The function \(f\) has a global minimum of \(2\) at \(x=0\) and is unbounded above, so its range is \([2, \infty)\). 2. The reciprocal \(g\) has a global maximum of \(\frac{1}{2}\) at \(x=0\), is always positive, and approaches \(0\) without reaching it. Its range is \(\left(0, \frac{1}{2}\right]\). 3. Since \(f(-x)=f(x)\), \(f\) is even. Then \(g(-x)=\frac{1}{f(-x)}=\frac{1}{f(x)}=g(x)\), so \(g\) is also even. 4. Both extrema occur at \(x=0\), but \(f\) has a minimum at \((0, 2)\), while \(g\) has a maximum at \(\left(0, \frac{1}{2}\right)\).

1) Range of \(f\): \([2, \infty)\); range of \(g\): \(\left(0, \frac{1}{2}\right]\) 2) Both graphs are symmetric about the y-axis. 3) \(f\) has a minimum at \((0, 2)\); \(g\) has a maximum at \(\left(0, \frac{1}{2}\right)\).

- Start with familiar power functions. - Consider how the sign of the leading coefficient affects the graph. - For part c, look for a function that is increasing across its entire domain.

1. A quadratic with a positive leading coefficient opens upward and has one local minimum. One example is \(h(x)=x^2\). 2. The fourth-degree polynomial \(h(x)=-x^4\) has exactly one local maximum, at the origin. 3. A strictly increasing function has no local extrema. One example is \(h(x)=x^3\).

Possible answers: a) \(h(x)=x^2\) b) \(h(x)=-x^4\) c) \(h(x)=x^3\)

- Locate the two turning points first. - Trace the curve from left to right and note where y-values rise or fall as \(x\) increases. - Use the turning-point x-values as interval boundaries.

1. The graph rises until the local maximum at \(x=-1\). 2. It falls from \(x=-1\) to the local minimum at \(x=1\). 3. It rises again for \(x>1\).

Increasing on \(( -\infty,-1)\) and \((1,\infty)\); decreasing on \((-1,1)\).

- Start with the maximum-turning-points relationship for a degree-\(n\) polynomial. - The observed number of turning points cannot exceed that maximum. - Check the smallest degree for which the maximum is at least \(4\).

1. A degree-\(n\) polynomial can have at most \(n-1\) turning points. 2. Four turning points require \(n-1\ge4\), so \(n\ge5\). 3. A degree-\(5\) polynomial can attain the maximum of \(4\) turning points. Therefore, the least possible degree is \(5\).

The least possible degree is \(5\), because a degree-\(n\) polynomial has at most \(n-1\) turning points.

- Identify the symmetry from the powers in the polynomial. - Reflect the positive x-coordinates across the y-axis. - Use the positive leading term and even degree to determine what happens on both ends. - Local extrema alternate in type along a smooth polynomial graph.

1. The polynomial contains only even powers, so it is even and its graph is symmetric about the y-axis. 2. Therefore, the extrema at \(x=1\) and \(x=2\) have matching extrema of the same type at \(x=-1\) and \(x=-2\). 3. The leading term is \(\frac{1}{6}x^6\), so \(f(x)\to\infty\) as \(x\to\pm\infty\). Thus, the outer extrema at \(x=\pm2\) are local minima. 4. The extrema alternate in type, so \(x=\pm1\) are local maxima and \(x=0\) is a local minimum.

a) \(x=-1\) and \(x=-2\) b) Local minima: \(x=-2, 0, 2\) Local maxima: \(x=-1, 1\)

- Identify the symmetry from the odd powers. - Reflect each extremum through the origin. - Determine what happens to a peak under a 180-degree rotation.

1. The polynomial contains only odd powers, so \(g\) is odd and its graph is symmetric about the origin. 2. The extrema at \(x=1\) and \(x=2\) are reflected to extrema at \(x=-1\) and \(x=-2\). 3. A 180-degree rotation about the origin changes a local maximum into a local minimum and a local minimum into a local maximum. 4. Therefore, \(x=-1\) is a local minimum, and \(x=-2\) is a local maximum.

a) \(x=-1\) and \(x=-2\) b) \(x=-1\) is a local minimum, and \(x=-2\) is a local maximum.

- Write the quadratic in vertex form. - A positive reciprocal is largest when its denominator is smallest. - What happens to \(\frac{1}{f(x)}\) when \(f(x)\) becomes very large?

1. Complete the square: \(f(x)=(x-3)^2+2\). Therefore, \(f\) has a global minimum at \((3, 2)\). 2. The reciprocal \(g(x)=\frac{1}{(x-3)^2+2}\) is largest when its positive denominator is smallest. Thus, \(g\) has a global maximum at \(\left(3, \frac{1}{2}\right)\). 3. As \(x\to\pm\infty\), \(f(x)\to\infty\), so \(g(x)\to 0\). The horizontal asymptote of \(g\) is \(y=0\). 4. More generally, for positive function values, taking reciprocals reverses order. A local minimum of \(f\) becomes a local maximum of \(\frac{1}{f}\), and a local maximum becomes a local minimum.

1) \(f\): global minimum at \((3, 2)\); \(g\): global maximum at \(\left(3, \frac{1}{2}\right)\) 2) \(f(x)\to\infty\) and \(g(x)\to 0\) as \(x\to\pm\infty\) 3) For positive \(f\), minima and maxima switch when taking the reciprocal.

- Rewrite the quadratic in vertex form. - Express both vertex coordinates in terms of \(k\). - Eliminate \(k\) and include the restriction implied by \(k>0\).

1. Rewrite the function in vertex form: \(f_k(x)=-\frac{2}{k}(x-2k)^2+8k+1\). 2. Thus, the vertex has coordinates \((2k,8k+1)\). 3. From \(x=2k\), obtain \(k=\frac{x}{2}\). Substitute into the y-coordinate: \(y=8\left(\frac{x}{2}\right)+1=4x+1\). 4. Because \(k>0\), the vertex satisfies \(x>0\). Therefore, the locus is \(y=4x+1\), for \(x>0\).

\(y=4x+1\), for \(x>0\)

- Local maxima and minima alternate. - Follow the only possible shape through the extrema from left to right. - Use the fact that the function is a polynomial and has no other extrema. - Count x-axis crossings from the end behavior and the positive extremum values.

1. Consecutive local extrema alternate between maxima and minima. Since the function value decreases from \(A\) to \(B\) and then increases from \(B\) to \(C\), \(A\) and \(C\) are local maxima, while \(B\) is a local minimum. 2. To the left of the leftmost local maximum, the polynomial must increase toward \(A\), so \(f(x)\to-\infty\) as \(x\to-\infty\). To the right of the rightmost local maximum, it must decrease away from \(C\), so \(f(x)\to-\infty\) as \(x\to\infty\). 3. All three extremum values are positive. The graph rises from \(-\infty\) to \(A\), so it crosses the x-axis exactly once before \(x=-2\). Between the extrema, the graph remains above the x-axis. After \(C\), it falls to \(-\infty\), so it crosses exactly once more. Therefore, \(f\) has exactly two real zeros.

a) \(A\) and \(C\) are local maxima; \(B\) is a local minimum. b) \(f(x)\to-\infty\) as \(x\to\pm\infty\). c) Exactly two real zeros.

- Substitute both given points into \(f(x) = ax^2 + bx + c\). - A downward-opening parabola has a negative leading coefficient. - Use the vertex formula \(x_v = -\frac{b}{2a}\) and require \(x_v > 0\).

1. Start with \(f(x) = ax^2 + bx + c\). Since the graph passes through \((0, 1.5)\), \(c = 1.5\). 2. Since the graph also passes through \((30, 0)\), \(900a + 30b + 1.5 = 0\). Solving for \(b\) gives \(b = -30a - 0.05\). Therefore, \(f_a(x) = ax^2 - (30a + 0.05)x + 1.5\). 3. The parabola opens downward when \(a < 0\). Its vertex has \(x\)-coordinate \(x_v = -\frac{b}{2a} = \frac{30a + 0.05}{2a}\). 4. Requiring \(x_v > 0\), while \(a < 0\), gives \(30a + 0.05 < 0\). Thus, \(a < -\frac{1}{600}\). 5. The maximum height occurs at the vertex, whose \(x\)-coordinate is \(x_v = \frac{30a + 0.05}{2a} = 15 + \frac{0.025}{a}\).

a) \(f_a(x) = ax^2 - (30a + 0.05)x + 1.5\) b) \(a < -\frac{1}{600}\) c) \(x = 15 + \frac{0.025}{a}\) feet from the nozzle

- Compare the two ends of an even-degree polynomial graph. - Think about what continuity forces when both ends rise or both ends fall. - Apply the stated degree-versus-turning-points relationship with \(n=4\).

1. For an even-degree polynomial, both ends rise when the leading coefficient is positive and both ends fall when the leading coefficient is negative. 2. Because polynomial graphs are continuous, a positive-leading even-degree polynomial must attain an absolute minimum, while a negative-leading even-degree polynomial must attain an absolute maximum. 3. A quartic has degree \(4\). Using the degree-turning-point bound, it can have at most \(4-1=3\) turning points.

a) A positive-leading even-degree polynomial has an absolute minimum; a negative-leading even-degree polynomial has an absolute maximum. b) A quartic can have at most \(3\) turning points because a degree-\(4\) polynomial can have at most \(4-1\) turning points.

- Identify each peak and valley. - Compare the y-values of all extrema. - Include the endpoint values when finding absolute extrema on a closed interval. - A local extremum need not be the highest or lowest point on the entire interval.

1. From the graph, the local maxima are \((-2.5, 3)\) and \((1, 4)\), and the local minima are \((-1, 1)\) and \((3, -3)\). 2. The endpoint values are \(f(-4)=-2\) and \(f(4)=-1\). The largest value on the interval is \(4\), so \((1, 4)\) is the absolute maximum. The smallest value is \(-3\), so \((3, -3)\) is the absolute minimum. The other two extrema are local only.

a) Local maxima: \((-2.5, 3)\), \((1, 4)\); local minima: \((-1, 1)\), \((3, -3)\) b) Absolute maximum: \((1, 4)\); absolute minimum: \((3, -3)\). The points \((-2.5, 3)\) and \((-1, 1)\) are local only.

- Use the sign of the leading term to determine end behavior. - Rewrite the function to identify its absolute extrema. - Compare the value at \(x=0\) and nearby values with the displayed graph.

1. The leading coefficient of \(k\) is negative and its degree is even, so \(k(x)\to-\infty\) as \(x\to\pm\infty\). The displayed graph rises at both ends. 2. Rewrite \(k(x)=2-(x^2-1)^2\). Therefore, \(k\) has absolute maxima of \(2\) at \(x=\pm1\), while the displayed graph has absolute minima. 3. Also, \(k(0)=1\), and nearby values such as \(k(0.5)=1.4375\) are larger, so \((0,1)\) is a local minimum. The displayed graph has a local maximum at \(x=0\) and y-intercept \(-1\).

1. \(k(x)\to-\infty\) as \(x\to\pm\infty\), but the displayed graph rises at both ends. 2. \(k\) has absolute maxima at \(x=\pm1\), while the graph has absolute minima. 3. \(k\) has a local minimum at \((0,1)\), while the graph has a local maximum at \((0,-1)\).

- Count the graph's turning points and relate that number to polynomial degree. - Decide whether the graph has y-axis symmetry or \(180^\circ\) rotational symmetry about the origin. - For origin symmetry, keep only odd powers in the cubic form. - Use both the coordinates and the horizontal tangent at the marked local maximum.

1. The graph has two local extrema. A polynomial of degree \(n\) has at most \(n-1\) turning points, so its degree is at least \(3\). A cubic is therefore the least possible degree. 2. The graph has \(180^\circ\) rotational symmetry about the origin, so \(f\) is odd. For a cubic, \(f(x)=ax^3+cx\). 3. The marked local maximum is \((-1, 2)\). Therefore, \(f(-1)=2\) and \(f'(-1)=0\). These conditions give \(-a-c=2\) and \(3a+c=0\). Solving the system gives \(a=1\) and \(c=-3\). Thus, \(f(x)=x^3-3x\).

1. Least possible degree: \(3\) 2. \(f(x)=ax^3+cx\) 3. \(f(x)=x^3-3x\)

- Count the graph's turning points and relate that number to polynomial degree. - Compare the left and right halves of the graph to identify its symmetry. - Determine which powers can appear in a polynomial that is symmetric about the y-axis. - Use both the coordinates and the horizontal tangent at one marked extremum.

1. The graph has three local extrema. A polynomial of degree \(n\) has at most \(n-1\) turning points, so the degree must be at least \(4\). 2. The graph is symmetric about the y-axis, so \(f\) is even. For a fourth-degree polynomial, write \(f(x)=ax^4+bx^2+c\). 3. The marked local maximum \((0, 0)\) gives \(c=0\), so \(f(x)=ax^4+bx^2\). The local minimum \((2, -4)\) gives \(16a+4b=-4\). Because the tangent there is horizontal, \(f'(2)=0\). Since \(f'(x)=4ax^3+2bx\), this gives \(32a+4b=0\). Solving the system yields \(a=\frac{1}{4}\) and \(b=-2\). Therefore, \(f(x)=\frac{1}{4}x^4-2x^2\).

1. The degree is at least \(4\) because the graph has three turning points. 2. The graph is symmetric about the y-axis, so \(f(x)=ax^4+bx^2+c\). 3. \(f(x)=\frac{1}{4}x^4-2x^2\)

- Interpret \(f(x)=c\) as intersections with a horizontal line \(y=c\). - Identify the y-values of the local maximum and the two global minima. - Check what happens when the horizontal line passes exactly through an extremum. - Consider the graph's end behavior beyond the visible turning points.

1. Solutions to \(f(x)=c\) are the x-coordinates where the horizontal line \(y=c\) intersects the graph. 2. For \(c=0\), the graph crosses the x-axis four times, so there are four solutions. 3. The graph has two global minima at height \(-4\) and a local maximum at height \(6\). A horizontal line intersects all four branches when \(-4<c<6\). 4. At \(c=-4\), the line is tangent at the two global minima, giving exactly two solutions. For \(c>6\), the line intersects only the two outer branches, also giving exactly two solutions. 5. At \(c=6\), the line passes through the local maximum and intersects both outer branches, giving exactly three solutions.

a) \(4\) solutions b) \(-4<c<6\) c) \(c=-4\) or \(c>6\) d) \(c=6\)

- Complete the square to identify the vertex. - Compare the vertex's y-coordinate with zero. - Substitute the proposed common point into the family equation.

1. Complete the square: \(g_k(x)=\left(x-\frac{k}{2}\right)^2-\frac{k^2}{4}+2k-3\). Therefore, the vertex is \(\left(\frac{k}{2},-\frac{k^2}{4}+2k-3\right)\). 2. Because each parabola opens upward, the number of real zeros depends on the vertex's y-coordinate. Factor it: \(-\frac{k^2}{4}+2k-3=-\frac{1}{4}(k-2)(k-6)\). 3. The vertex is below the x-axis when \(k<2\) or \(k>6\), on the x-axis when \(k=2\) or \(k=6\), and above it when \(2<k<6\). Therefore, the graph has two, one, or no real zeros respectively. 4. Substitute \(x=2\): \(g_k(2)=4-2k+2k-3=1\). Thus, \((2,1)\) lies on every graph.

a) \(g_k(x)=\left(x-\frac{k}{2}\right)^2-\frac{k^2}{4}+2k-3\); vertex \(\left(\frac{k}{2},-\frac{k^2}{4}+2k-3\right)\) b) Two real zeros for \(k<2\) or \(k>6\); one real zero for \(k=2\) or \(k=6\); no real zeros for \(2<k<6\) c) \(g_k(2)=1\) for every \(k\).

- Complete the square to find the vertex. - Eliminate the parameter from the vertex coordinates. - Compare the vertex's y-coordinate with zero to count real zeros. - Collect the terms containing \(a\) to find where the function value is independent of the parameter.

1. Complete the square: \(f_a(x)=(x-a)^2-a^2+3a\). Therefore, the vertex is \((a,-a^2+3a)\). 2. At a vertex, \(x=a\). Replacing \(a\) with \(x\) in the y-coordinate gives the locus \(y=-x^2+3x\). 3. Each parabola opens upward, so the number of real zeros depends on \(-a^2+3a=a(3-a)\). It is negative for \(a<0\) or \(a>3\), zero for \(a=0\) or \(a=3\), and positive for \(0<a<3\). 4. Thus, there are two real zeros for \(a<0\) or \(a>3\), one real zero for \(a=0\) or \(a=3\), and no real zeros for \(0<a<3\). 5. Rewrite \(f_a(x)=x^2+a(3-2x)\). For the value to be independent of \(a\), require \(3-2x=0\), so \(x=\frac{3}{2}\). Then \(y=\frac{9}{4}\). The common point is \(\left(\frac{3}{2},\frac{9}{4}\right)\).

a) \(f_a(x)=(x-a)^2-a^2+3a\); vertex \((a,-a^2+3a)\) b) \(y=-x^2+3x\) c) Two real zeros for \(a<0\) or \(a>3\); one for \(a=0\) or \(a=3\); none for \(0<a<3\) d) \(\left(\frac{3}{2},\frac{9}{4}\right)\)

- Complete the square to identify the vertex. - Eliminate \(t\) from the vertex coordinates. - When does the vertex of a downward-opening parabola lie on the x-axis? - Collect the terms containing \(t\) to find a parameter-independent point.

1. Complete the square: \(h_t(x)=-\left(x-\frac{t}{2}\right)^2+\frac{(t-4)^2}{4}\). 2. Therefore, the vertex is \(\left(\frac{t}{2},\frac{(t-4)^2}{4}\right)\). 3. From \(x=\frac{t}{2}\), obtain \(t=2x\). Substitution gives the vertex locus \(y=(x-2)^2\). 4. A downward-opening parabola is tangent to the x-axis when its vertex lies on the axis. Thus, \(\frac{(t-4)^2}{4}=0\), so \(t=4\). 5. Rewrite \(h_t(x)=-x^2+4+t(x-2)\). At \(x=2\), the parameter term is zero and \(h_t(2)=0\). Therefore, every graph passes through \((2,0)\).

a) \(h_t(x)=-\left(x-\frac{t}{2}\right)^2+\frac{(t-4)^2}{4}\); vertex \(\left(\frac{t}{2},\frac{(t-4)^2}{4}\right)\) b) \(y=(x-2)^2\) c) \(t=4\) d) \((2,0)\)

- Use the vertex formula for a quadratic. - Express the vertex coordinates in terms of \(a\), then eliminate the parameter. - Determine when the parabola opens downward. - Account for the sign of \(a\) when solving the inequality for the vertex's height.

1. The vertex formula gives \(x_v=\frac{2}{a}\). 2. Evaluate the function at the vertex: \(y_v=f_a\left(\frac{2}{a}\right)=a-\frac{4}{a}\). 3. From \(x=\frac{2}{a}\), obtain \(a=\frac{2}{x}\). Substituting gives \(y=\frac{2}{x}-2x\), with \(x\ne0\). 4. The vertex is a maximum when the parabola opens downward, so \(a<0\). 5. It lies below the x-axis when \(a-\frac{4}{a}<0\). Since \(a<0\), multiplying by \(a\) reverses the inequality: \(a^2-4>0\). Combined with \(a<0\), this gives \(a<-2\).

a) \(y=\frac{2}{x}-2x\), for \(x\ne0\) b) \(a<-2\)

- Interpret \(g(x)=k\) as intersections with a horizontal line \(y=k\). - Track what changes when the line passes through the minimum, the x-axis, and the maximum. - Remember that the right branch approaches \(y=0\) but does not reach it again.

1. Zeros are x-intercepts. The graph crosses the x-axis at \(x=-3\) and \(x=0\), so \(g\) has two zeros. 2. A horizontal line \(y=k\) intersects the graph twice when \(-2<k\le0\). It also intersects twice when \(k=2\): once on the left branch and once at the local maximum. 3. When \(0<k<2\), the horizontal line intersects the left branch once and the right-hand hump twice, for three solutions. 4. When \(k<-2\), the horizontal line lies below the global minimum and does not intersect the graph. At \(k=-2\), there is one solution.

a) \(2\) b) \(k\in(-2, 0]\cup\{2\}\) c) \(k\in(0, 2)\) d) \(k<-2\)