Free math worksheets

- Function notation tells you which input replaces \(x\). - Use parentheses when substituting a negative input. - Multiply before adding.

1. Substitute \(x=2\): \(f(2)=0.4(2)+1.2=2\). 2. Substitute \(x=-3\): \(f(-3)=0.4(-3)+1.2=0\).

1) \(f(2)=2\) 2) \(f(-3)=0\)

- Substitute the given input into the function. - Apply the exponent before multiplication and addition. - The negative coefficient is outside the square.

1. Substitute \(x = 2\): \(h(2) = -0.5 \cdot 2^2 + 2 \cdot 2 + 1 = -2 + 4 + 1 = 3\). The height is \(3\,\text{m}\). 2. Substitute \(x = 4\): \(h(4) = -0.5 \cdot 4^2 + 2 \cdot 4 + 1 = -8 + 8 + 1 = 1\). The height is \(1\,\text{m}\).

1) \(3\,\text{m}\) 2) \(1\,\text{m}\)

- In a function table, match each input in the top row with the output directly below it. - In \(f(a)=b\), \(a\) is the input and \(b\) is the output. - An input-output pair can be written as an ordered pair \((x,y)\).

1. In the column where \(x=2\), the output is \(-3\), so \(f(2)=-3\). 2. In the column where \(x=-3\), the output is \(7\), so \(f(-3)=7\). 3. The statement \(f(5)=-9\) means input \(5\) has output \(-9\), so the ordered pair is \((5,-9)\).

a) \(-3\) b) \(7\) c) \((5,-9)\)

- What is the \(x\)-coordinate of every point on the \(y\)-axis? - Substitute that input into each function. - Check whether any substitution creates division by zero.

1. Substitute \(x = 0\) into \(f\): \(f(0) = 4(0) - 2 = -2\). Therefore, the graph of \(f\) crosses the \(y\)-axis at \((0, -2)\). 2. Substituting \(x = 0\) into \(g\) would give \(g(0) = \frac{8}{0}\), which is undefined. Therefore, \(0\) is not in the domain of \(g\), and its graph does not cross the \(y\)-axis. 3. Substitute \(x = 0\) into \(h\): \(h(0) = \frac{0}{2} = 0\). Therefore, the graph of \(h\) crosses the \(y\)-axis at \((0, 0)\).

\(f\): The graph crosses the \(y\)-axis at \((0, -2)\). \(g\): The graph does not cross the \(y\)-axis because \(g(0)\) is undefined. \(h\): The graph crosses the \(y\)-axis at \((0, 0)\).

- Use parentheses when substituting a negative input. - Evaluate exponents before multiplication and addition. - Track the sign of each term.

1. \(T(-3)=(-3)^2-3\cdot(-3)+1=9+9+1=19\). 2. \(T(0.5)=(0.5)^2-3\cdot0.5+1=0.25-1.5+1=-0.25\). 3. \(T(2)=2^2-3\cdot2+1=4-6+1=-1\).

\(T(-3)=19\) \(T(0.5)=-0.25\) \(T(2)=-1\)

- Substitute each listed input for \(x\) in the function rule. - Keep the input and its output paired as you work. - For part b), compare the outputs after you have evaluated all five inputs.

1. Substitute each input into \(T(x)=x^2+5\): \(T(1)=6\), \(T(2)=9\), \(T(3)=14\), \(T(4)=21\), and \(T(5)=30\). 2. Compare the five outputs. The greatest is \(30\), which occurs at input \(x=5\).

a) \(T(1)=6\), \(T(2)=9\), \(T(3)=14\), \(T(4)=21\), and \(T(5)=30\) b) \(x=5\)

- Evaluate the function at each point's x-coordinate. - Compare the calculated function value with the point's y-coordinate. - What does it mean when the point's y-coordinate is greater than or less than the function value?

1. For \(P_1\), \(f(2)=2^2-3.5=0.5\). This equals the point's y-coordinate, so \(P_1\) lies on the graph. 2. For \(P_2\), \(f(-1.5)=(-1.5)^2-3.5=2.25-3.5=-1.25\). This equals the point's y-coordinate, so \(P_2\) lies on the graph. 3. For \(P_3\), \(f(0.5)=(0.5)^2-3.5=0.25-3.5=-3.25\). Since \(-3.75<-3.25\), \(P_3\) lies below the graph. 4. For \(P_4\), \(f(-1)=(-1)^2-3.5=1-3.5=-2.5\). Since \(1>-2.5\), \(P_4\) lies above the graph.

\(P_1\): on the graph \(P_2\): on the graph \(P_3\): below the graph \(P_4\): above the graph

- Substitute the point's x-coordinate into the function and compare the result with its y-coordinate. - For part b), first isolate the squared expression before considering which real inputs can produce the target output. - Function notation \(f(-3)\) means to substitute \(-3\) for \(x\).

1. Evaluate the function at \(x=-2\): \(f(-2)=\frac{1}{4}(-2)^2=1\). Therefore, \(P\) lies on the graph. 2. Solve \(\frac{1}{4}x^2=16\). Multiplying by \(4\) gives \(x^2=64\), so \(x=-8\) or \(x=8\). 3. \(f(-3)=\frac{1}{4}(-3)^2=\frac{9}{4}=2.25\).

a) Yes, because \(f(-2)=1\). b) \(x=-8\) and \(x=8\) c) \(2.25\)

- Read the function name carefully before substituting the requested input. - Use parentheses around a negative input before evaluating powers. - Evaluate powers before combining the remaining terms.

1. \(f(-1)=5\cdot(-1)^4-3\cdot(-1)^3+2\cdot(-1)^2-(-1)+4=15\). 2. \(g\left(-\frac{1}{2}\right)=8\cdot\left(-\frac{1}{2}\right)^3+4\cdot\left(-\frac{1}{2}\right)^2-2\cdot\left(-\frac{1}{2}\right)+1=2\).

a) \(15\) b) \(2\)

- Match each requested value with the correct function rule. - Substitute the input before evaluating powers and products. - In part c), evaluate each factor before multiplying.

1. \(f(5)=4\cdot5^2-3\cdot5+12=97\). 2. \(g(3)=2\cdot3^3+5\cdot3-8=61\). 3. \(h(6)=(6+4)(2\cdot6-1)=110\).

a) \(97\) b) \(61\) c) \(110\)

- Cube the input before multiplying by \(6\). - The exponent applies only to \(n\). - Calculate carefully with the decimal input.

1. \(T(10)=6\cdot10^3+4\cdot10+7=6000+40+7=6047\). 2. \(T(2)=6\cdot2^3+4\cdot2+7=48+8+7=63\). 3. \(T(0.5)=6\cdot0.5^3+4\cdot0.5+7=0.75+2+7=9.75\).

a) \(6047\) b) \(63\) c) \(9.75\)

- A point \((x, y)\) lies on a function's graph when substituting its x-coordinate gives its y-coordinate. - Substitute each x-coordinate into \(f(x)\). - Pay close attention to parentheses and signs when evaluating powers of negative numbers.

1. For \(A\), \(f(3)=3^3-4\cdot 3=27-12=15\), which matches the y-coordinate. Therefore, \(A\) lies on the graph. 2. For \(B\), \(f(-2)=(-2)^3-4\cdot(-2)=-8+8=0\), which matches the y-coordinate. Therefore, \(B\) lies on the graph. 3. For \(C\), \(f(1)=1^3-4\cdot 1=-3\), which does not equal \(3\). Therefore, \(C\) does not lie on the graph. 4. For \(D\), \(f(-1)=(-1)^3-4\cdot(-1)=-1+4=3\), which matches the y-coordinate. Therefore, \(D\) lies on the graph.

Points \(A\), \(B\), and \(D\) lie on the graph. Point \(C\) does not.

- A point on the graph must satisfy \(y=x^5\). - When the output is known and the input is missing, solve the fifth-power equation for the input. - Track the sign of a negative number raised to an odd power. - You may rewrite decimals as fractions if that makes a power easier to evaluate.

1. For \(A\), \(y_A=3^5=243\), so \(A(3,243)\). 2. For \(B\), solve \(x_B^5=-32\). Thus, \(x_B=\sqrt[5]{-32}=-2\), so \(B(-2,-32)\). 3. For \(C\), \(y_C=(-0.1)^5=-0.00001\), so \(C(-0.1,-0.00001)\). 4. For \(D\), \(x_D^5=0\), so \(x_D=0\) and \(D(0,0)\). 5. For \(E\), \(y_E=(0.5)^5=0.03125\), so \(E(0.5,0.03125)\).

a) \(A(3,243)\) b) \(B(-2,-32)\) c) \(C(-0.1,-0.00001)\) d) \(D(0,0)\) e) \(E(0.5,0.03125)\)

- A relation defines \(y\) as a function of \(x\) when each allowed x-value corresponds to exactly one y-value. - Try solving each equation for \(y\). - A single x-value that produces two different y-values is enough to show that a relation is not a function.

1. Equation (1) gives \(y^3=27-x\), so \(y=\sqrt[3]{27-x}\). Every real x-value produces exactly one real y-value, so it defines a function. 2. For equation (2), choose \(x=0\). Then \(y^2=16\), so \(y=4\) or \(y=-4\). One x-value has two y-values, so the equation does not define \(y\) as a function of \(x\). 3. Equation (3) gives \(y=|x|+2\). Every real x-value produces exactly one y-value, so it defines a function.

(1) Function (2) Not a function (3) Function

- A relation defines \(y\) as a function of \(x\) only when each allowed x-value corresponds to exactly one y-value. - Try isolating \(y\). - Equations involving \(y^2\) or \(|y|\) may produce two y-values. - Check values that make the coefficient of \(y\) equal to \(0\).

1. For a), solving for \(y\) gives \(y=x^2+5\). Each x-value gives exactly one y-value, so this is a function. 2. For b), choose \(x=3\). Then \(y^2=4\), so \(y=2\) or \(y=-2\). One x-value has two y-values, so this is not a function. 3. For c), the equation reduces to \(-2x=6\), so \(x=-3\). When \(x=-3\), every real value of \(y\) satisfies the equation. Therefore, this is not a function. 4. For d), choose \(x=0\). Then \(|y|=3\), so \(y=3\) or \(y=-3\). Therefore, this is not a function. 5. For e), choose \(x=2\). Then \(0y=0\), which is true for every real \(y\). Therefore, this is not a function.

a) Function: \(y=x^2+5\) b) Not a function c) Not a function d) Not a function e) Not a function

- Use the definition: each input may have only one output. - A single input with two possible outputs is enough to disprove that a relation is a function. - For an equation already solved for \(y\), ask whether the right-hand side produces one value or more than one value for each input.

1. For a), choose \(x=4\). Both \(y=2\) and \(y=-2\) satisfy the relation, so one input has two outputs. It is not a function. 2. In b), the explicit formula gives exactly one value of \(y\) for every real \(x\), so it is a function. 3. In c), the expression \(2^x\) gives exactly one positive value for every real \(x\), so it is a function. 4. In d), the polynomial expression gives exactly one real output for every real input, so it is a function.

a) Not a function; for example, \(x=4\) gives \(y=2\) and \(y=-2\). b) Function; each real input gives one output. c) Function; each real input gives one output. d) Function; each real input gives one output.

- Start at the requested x-value and move vertically until you reach the graph. - The y-coordinate of that point is the function value. - Keep the same input when you write the corresponding ordered pair.

1. At \(x=-2\), the graph has y-coordinate \(5\), so \(f(-2)=5\). 2. At \(x=4\), the graph has y-coordinate \(-1\), so \(f(4)=-1\). 3. Therefore, the corresponding point is \((4,-1)\).

a) \(5\) b) \(-1\) c) \((4,-1)\)

- Read \(C(h)\) as the cost associated with an input of \(h\) hours. - For part b, substitute the requested number of hours for the input variable. - Use the context, not just the formula, to attach units to each quantity.

1. The statement \(C(3)=11.50\) means that parking for \(3\) hours costs \(\$11.50\). 2. Evaluate \(C(6)=4+2.5(6)=19\), so parking for \(6\) hours costs \(\$19\). 3. The input is time in hours, and the output is cost in dollars.

a) Parking for \(3\) hours costs \(\$11.50\). b) \(C(6)=19\), so parking for \(6\) hours costs \(\$19\). c) Input: hours; output: dollars

- Evaluate the function at each listed input before comparing outputs. - Compare inputs that have the same absolute value. - What happens to the sign when a number is squared?

1. Evaluate the function: \(T(-2)=1.5\), \(T(-1.5)=-1.125\), \(T(-0.5)=-4.125\), \(T(0.5)=-4.125\), \(T(1.5)=-1.125\), and \(T(2)=1.5\). 2. The equal-output pairs are \(-2\) and \(2\), \(-1.5\) and \(1.5\), and \(-0.5\) and \(0.5\). 3. Opposite inputs have the same square, so \(T(-x)=1.5(-x)^2-4.5=1.5x^2-4.5=T(x)\).

a) \(T(-2)=1.5\), \(T(-1.5)=-1.125\), \(T(-0.5)=-4.125\), \(T(0.5)=-4.125\), \(T(1.5)=-1.125\), and \(T(2)=1.5\) b) The pairs are \((-2,2)\), \((-1.5,1.5)\), and \((-0.5,0.5)\). Opposite inputs give equal outputs because \((-x)^2=x^2\).

- For a rational expression, evaluate the numerator and denominator separately. - Apply exponents before multiplication and subtraction. - Simplify each final fraction.

1. \(T(4)=\frac{4^2+1}{4+2}=\frac{17}{6}=2\frac{5}{6}\). 2. \(T\left(\frac{1}{2}\right)=\frac{\left(\frac{1}{2}\right)^2+1}{\frac{1}{2}+2}=\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{1}{2}\). 3. \(S(3)=3^3-2\cdot3^2+5=27-18+5=14\). 4. \(S(10)=10^3-2\cdot10^2+5=1000-200+5=805\).

a) \(T(4)=2\frac{5}{6}\); \(T\left(\frac{1}{2}\right)=\frac{1}{2}\) b) \(S(3)=14\); \(S(10)=805\)

- Evaluate the pieces of the numerator for each input before forming the quotient. - Keep each intermediate value in the same column as its input. - Round only the final function value in each column.

1. Square each input to obtain \(0.04\), \(0.16\), \(0.36\), \(0.64\), \(1.00\), and \(1.44\). 2. Double each input to obtain \(0.4\), \(0.8\), \(1.2\), \(1.6\), \(2.0\), and \(2.4\). 3. Substitute each input into \(f(x)=\frac{x^2+2x}{x+1.5}\) and round only the final quotient: \(0.26\), \(0.51\), \(0.74\), \(0.97\), \(1.20\), and \(1.42\).

\(x\)	\(0.2\)	\(0.4\)	\(0.6\)	\(0.8\)	\(1.0\)	\(1.2\)
\(x^2\)	\(0.04\)	\(0.16\)	\(0.36\)	\(0.64\)	\(1.00\)	\(1.44\)
\(2x\)	\(0.4\)	\(0.8\)	\(1.2\)	\(1.6\)	\(2.0\)	\(2.4\)
\(f(x)\)	\(0.26\)	\(0.51\)	\(0.74\)	\(0.97\)	\(1.20\)	\(1.42\)

- Replace the input variable with the entire expression inside the function parentheses. - Use parentheses when multiplying or squaring a complete function expression. - An identity must hold for every allowed value, not just one value.

1. \(f(2a)=8a-3\), while \(2f(a)=2(4a-3)=8a-6\). The expressions are not identical, so a is false. 2. \(f(x+1)=4(x+1)-3=4x+1\), and \(f(x)+4=(4x-3)+4=4x+1\). The expressions are identical, so b is true. 3. \(f(x^2)=4x^2-3\), while \((f(x))^2=(4x-3)^2=16x^2-24x+9\). These expressions are not identical, so c is false. They are equal only when \(x=1\).

a) False. \(f(2a)=8a-3\), while \(2f(a)=8a-6\), so the two sides are not identical. b) True. \(f(x+1)=4x+1\) and \(f(x)+4=4x+1\). c) False. \(f(x^2)=4x^2-3\), while \((f(x))^2=(4x-3)^2\), so the expressions are not identical.

- Even powers make negative inputs produce nonnegative outputs. - Use the power of a power property. - Relate \((\sqrt{2})^2\) to \(2\). - When substituting \(2a\), raise the entire product to the fourth power.

1. \(f(0.5)=(0.5)^4=0.0625\) and \(g(0.5)=(0.5)^6=0.015625\). 2. \(f(2)=2^4=16\) and \(g(2)=2^6=64\). 3. \(f(-3)=(-3)^4=81\) and \(g(-1.5)=(-1.5)^6=11.390625\). 4. \(g(\sqrt{2})=(\sqrt{2})^6=((\sqrt{2})^2)^3=2^3=8\). 5. \(f(2a)=(2a)^4=16a^4\), and \(g(a^2)=(a^2)^6=a^{12}\).

a) \(f(0.5)=0.0625\), \(g(0.5)=0.015625\), \(f(2)=16\), and \(g(2)=64\) b) \(f(-3)=81\) and \(g(-1.5)=11.390625\) c) \(g(\sqrt{2})=8\) d) \(f(2a)=16a^4\) and \(g(a^2)=a^{12}\)

- An odd power preserves the sign of the input. - What happens when a fifth root is raised to the fifth power? - Apply the outer exponent to every factor in \((a^3b)^5\). - Compare the values at \(x\) and \(-x\).

1. \(f(-2)=-32\), \(f(-1)=-1\), \(f(0)=0\), \(f(1)=1\), and \(f(2)=32\). 2. \(f(0.2)=(0.2)^5=0.00032\), and \(f(-0.4)=(-0.4)^5=-0.01024\). 3. \(f(\sqrt[5]{12})=(\sqrt[5]{12})^5=12\). 4. \(f(-a)=(-a)^5=-a^5\), and \(f(a^3b)=(a^3b)^5=a^{15}b^5\).

a) \(f(-2)=-32\), \(f(-1)=-1\), \(f(0)=0\), \(f(1)=1\), and \(f(2)=32\) b) \(f(0.2)=0.00032\) and \(f(-0.4)=-0.01024\) c) \(f(\sqrt[5]{12})=12\) d) \(f(-a)=-a^5\) and \(f(a^3b)=a^{15}b^5\)

- Substitute the coordinates of \(P\) into the equation to create an equation in \(a\). - After finding \(a\), evaluate the function at the x-coordinate of \(Q\). - An even power of a negative number is positive.

1. Substitute \(P(2, 4)\): \(4=a\cdot 2^4-2\cdot 2^2\). 2. Simplify: \(4=16a-8\), so \(12=16a\) and \(a=\frac{3}{4}=0.75\). 3. The function is \(f(x)=0.75x^4-2x^2\). 4. Evaluate at \(x=-1\): \(f(-1)=0.75(-1)^4-2(-1)^2=0.75-2=-1.25\). 5. This equals the y-coordinate of \(Q\), so \(Q\) lies on the graph.

a) \(a=0.75\) b) Yes. \(Q(-1, -1.25)\) lies on the graph.

- An even power gives the same output for opposite nonzero inputs. - When a sixth power equals a positive number, check whether both a positive and a negative input work. - Rewrite \(1{,}000{,}000\) as a power of \(10\). - Consider separately what happens when the output is zero.

1. For \(P_1\), \(y_1=(-2)^6=64\). 2. For \(P_2\), solve \(x_2^6=1\). The real solutions are \(x_2=1\) and \(x_2=-1\). 3. For \(P_3\), \(y_3=(0.5)^6=\frac{1}{64}=0.015625\). 4. For \(P_4\), solve \(x_4^6=1{,}000{,}000=10^6\). The real solutions are \(x_4=10\) and \(x_4=-10\). 5. For \(P_5\), \(x_5^6=0\), so \(x_5=0\).

a) \(P_1(-2,64)\) b) \(P_2(1,1)\) or \(P_2(-1,1)\) c) \(P_3(0.5,0.015625)\) d) \(P_4(10,1{,}000{,}000)\) or \(P_4(-10,1{,}000{,}000)\) e) \(P_5(0,0)\)

- Look for an x-value that makes more than one y-value possible. - In equation (1), consider when the product \(xy\) can equal zero without restricting \(y\) to one value. - Remember that \(\sqrt{a}\) denotes one nonnegative value, not both square roots. - In equation (2), consider whether opposite y-values can produce the same x-value.

1. In equation (1), let \(x=0\). Then \(0y=0\) is true for every real value of \(y\). Because one x-value corresponds to infinitely many y-values, the equation does not define a function. 2. In equation (2), let \(x=0.5\). Then \(0.5=\frac{1}{y^2+1}\), so \(y^2+1=2\) and \(y=\pm1\). Because one x-value corresponds to two y-values, the equation does not define a function. 3. In equation (3), the square-root symbol denotes the unique nonnegative square root. For every \(x\ge1\), the expression \(\sqrt{x-1}+2\) produces exactly one y-value, so it defines a function.

(1) Not a function (2) Not a function (3) Function

- Substitute a clearly readable point from each curve into the quadratic model. - Keep the scale factor \(\frac{v}{10}\) together before squaring. - For the percent comparison, identify which curve supplies the reference value.

1. Curve \(f\) contains the point \((20,8)\). Substituting gives \(8=k\left(\frac{20}{10}\right)^2=4k\), so \(k_f=2\). 2. Curve \(g\) contains the point \((20,4)\). Substituting gives \(4=4k\), so \(k_g=1\). 3. At \(v=25\), \(F_f=2(2.5)^2=12.5\,\text{kN}\), and \(F_g=(2.5)^2=6.25\,\text{kN}\). 4. The percent increase from \(g\) to \(f\) is \(\frac{12.5-6.25}{6.25}\cdot100\%=100\%\).

a) Curve \(f\): \(k=2\); curve \(g\): \(k=1\) b) The force on curve \(f\) is \(100\%\) greater.

- Evaluate both equations at the same input. - Keep all displayed digits through the first calculation. - Compare the final predictions by subtraction.

1. Using the full coefficients gives \(\hat y=1.2746\cdot 85+3.918=112.259\). 2. Using the rounded equation gives \(\hat y=1.3\cdot 85+3.9=114.4\). 3. The early rounding changes the prediction by \(114.4-112.259=2.141\).

Full-coefficient prediction: \(112.259\) Prematurely rounded prediction: \(114.4\) Rounding error: \(2.141\)

- Reverse function notation asks you to start from an output and locate the input or inputs that produce it. - Read across the output row first, then trace matching entries back to their inputs. - The definition of a function restricts the number of outputs for one input, not the number of inputs that may share one output.

1. In the output row, \(6\) occurs under inputs \(-4\) and \(2\), so \(q(-4)=6\) and \(q(2)=6\). 2. The statement \(q(5)=-3\) corresponds to the ordered pair \((5, -3)\). 3. A function requires each input to have exactly one output. Different inputs are allowed to share the same output, so the repeated output \(6\) does not violate the function definition.

a) \(a=-4\) and \(a=2\) b) \((5, -3)\) c) A function may assign the same output to different inputs; each individual input still has exactly one output.