Free Math Worksheets

- Replace \(x\) in the function rule with the given input. - Use parentheses when substituting a negative number. - Multiply before adding.

1. Substitute \(x=2\): \(y=0.4(2)+1.2=0.8+1.2=2\). 2. Substitute \(x=-3\): \(y=0.4(-3)+1.2=-1.2+1.2=0\).

1) \(y=2\) 2) \(y=0\)

- 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) \(1\,\text{m}\)

- 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. \(T(2)=2^2-3\cdot2+1=4-6+1=-1\).

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

- 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. - An equation of the form \(x^n=c\) can have two real solutions when \(n\) is even. - Function notation \(f(-3)\) means to substitute \(-3\) for \(x\).

1. Evaluate the function at \(x=-2\): \(f(-2)=\frac{1}{4}(-2)^4=\frac{1}{4}\cdot16=4\). Therefore, \(P\) lies on the graph. 2. Solve \(\frac{1}{4}x^4=64\). Multiplying by \(4\) gives \(x^4=256\), so \(x=4\) or \(x=-4\). 3. \(f(-3)=\frac{1}{4}(-3)^4=\frac{81}{4}=20.25\).

a) Yes, because \(f(-2)=4\). b) \(x=-4\) and \(x=4\) c) \(20.25\)

- Substitute the value using parentheses. - Evaluate all powers first. - An even power of a negative number is positive; an odd power is negative.

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

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

- Substitute the given value for the variable. - Evaluate exponents before multiplication and addition. - Evaluate each factor before multiplying in part c).

1. \(4\cdot5^2-3\cdot5+12=100-15+12=97\). 2. \(2\cdot3^3+5\cdot3-8=54+15-8=61\). 3. \((6+4)\cdot(2\cdot6-1)=10\cdot11=110\).

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 \(y=x^5\). - Use a fifth root when the output is known and the input is missing. - Track the sign of a negative number raised to an odd power. - You may rewrite decimals as fractions to simplify computation.

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

- 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) \(S(3)=14\); \(S(10)=805\)

- Calculate the parts of the numerator before finding the quotient. - Add \(1.5\) to each input for the denominator. - Round only the final value of \(y\).

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 the function and round only the final quotient: \(0.26,0.51,0.74,0.97,1.20,\) and \(1.42\).

<table> <tbody> <tr><td>\(x\)</td><td>\(0.2\)</td><td>\(0.4\)</td><td>\(0.6\)</td><td>\(0.8\)</td><td>\(1.0\)</td><td>\(1.2\)</td></tr> <tr><td>\(x^2\)</td><td>\(0.04\)</td><td>\(0.16\)</td><td>\(0.36\)</td><td>\(0.64\)</td><td>\(1.00\)</td><td>\(1.44\)</td></tr> <tr><td>\(2x\)</td><td>\(0.4\)</td><td>\(0.8\)</td><td>\(1.2\)</td><td>\(1.6\)</td><td>\(2.0\)</td><td>\(2.4\)</td></tr> <tr><td>\(y\)</td><td>\(0.26\)</td><td>\(0.51\)</td><td>\(0.74\)</td><td>\(0.97\)</td><td>\(1.20\)</td><td>\(1.42\)</td></tr> </tbody> </table>

- 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 b) True c) False

- 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. - If \(x^6\) equals a positive number, consider both the positive and negative sixth roots. - Rewrite \(1{,}000{,}000\) as a power of \(10\). - What input makes a sixth power equal to zero?

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

a) \(P_1(-2, 64)\) b) \(P_2(1, 1)\) or \(P_2(-1, 1)\) c) \(P_3(0.5, 0.015625)\) d) \(P_4(10, 1000000)\) or \(P_4(-10, 1000000)\) e) \(P_5(0, 0)\)

- Look for an x-value that makes more than one y-value possible. - In equation (1), test \(x=0\). - 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

- A function assigns exactly one output to each input in its domain. - Try a positive x-value in part a) and check whether two y-values work. - Identify each function family from the structure of its formula.

1. For a), choose \(x=4\). Both \(y=2\) and \(y=-2\) satisfy \(y^2=4\), so one input has two outputs. This relation is not a function. 2. The formula \(f(x)=-1.5x+4\) assigns exactly one output to every real input. It is a linear function. 3. The formula \(h(x)=2^x\) assigns exactly one positive output to every real input. It is an exponential function. 4. The formula \(k(x)=x^4-2x^2+1\) assigns exactly one output to every real input. It is a polynomial function of degree \(4\).

a) Not a function b) Function; linear c) Function; exponential d) Function; polynomial of degree \(4\)

- Substitute a clearly readable point from each curve into the model. - Divide the speed by \(10\) before squaring. - Use curve B as the reference value for the percent increase.

1. Curve A contains the point \((20, 8)\). Substituting gives \(8 = k\left(\frac{20}{10}\right)^2 = 4k\), so \(k_A = 2\). 2. Curve B contains the point \((20, 4)\). Substituting gives \(4 = 4k\), so \(k_B = 1\). 3. At \(v = 25\), \(F_A = 2(2.5)^2 = 12.5\,\text{kN}\), and \(F_B = (2.5)^2 = 6.25\,\text{kN}\). 4. The percent increase from B to A is \(\frac{12.5 - 6.25}{6.25} \cdot 100\% = 100\%\).

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