1.
Question 1. Find the equation of a line passing through the points \((-1,5)\) and \((3,17)\).
Solution:We first find the slope:\[m=\frac{y_2-y_1}{x_2-x_1}=\frac{17-5}{3-(-1)}=\frac{12}{4}=3\]Use point-slope form with \((-1,5)\):\[y-5=3(x-(-1))\]\[y-5=3x+3\]\[y=3x+8\]Answer: C
2.
Question 2. What is the degree of the equation \(f(x)=5x-4x^2-x^3\)?
Solution:The degree of a polynomial is the highest power of \(x\).In \(f(x)=5x-4x^2-x^3\), the highest exponent is \(3\).Answer: D
3.
Question 3. What do all members of the family of lines of the form \(y=2x+b\) have in common?
Solution:The slope-intercept form is \(y=mx+b\), where \(m\) is the slope.For \(y=2x+b\), the coefficient of \(x\) is \(2\).Therefore, every line in this family has slope \(2\).Answer: E
4.
Question 4. The functions \(f(x)=\frac{1}{x^3}\) and \(g(x)=x^{-\frac{1}{3}}\) are inverse of each other.
Solution:To check, find the inverse of \(y=\frac{1}{x^3}\).\[x=\frac{1}{y^3}\]\[y^3=\frac{1}{x}\]\[y=\sqrt[3]{\frac{1}{x}}=x^{-\frac{1}{3}}\]This matches \(g(x)=x^{-\frac{1}{3}}\), so the functions are inverses.Answer: A
5.
Question 5. Find the inverse of \(f(x)=x^7\).
Solution:Let \(y=x^7\).Switch \(x\) and \(y\): \(x=y^7\).Solve for \(y\):\[y=\sqrt[7]{x}=x^{\frac{1}{7}}\]Answer: C
6.
Question 6. Evaluate \(\log_5\left((-5)^2\right)\).
Solution:First simplify inside the logarithm:\[(-5)^2=25\]Then evaluate:\[\log_5(25)=\log_5(5^2)=2\]Answer: B
7.
Question 7. Rewrite the expression \(-\frac{1}{2}\log(x)+2\log(3)\) as a single logarithm.
Solution:Use the power rule of logarithms:\[-\frac{1}{2}\log(x)=\log\left(x^{-\frac{1}{2}}\right)=\log\left(\frac{1}{\sqrt{x}}\right)\]\[2\log(3)=\log(3^2)=\log(9)\]Combine using \(\log(a)+\log(b)=\log(ab)\):\[\log\left(\frac{1}{\sqrt{x}}\right)+\log(9)=\log\left(\frac{9}{\sqrt{x}}\right)\]Answer: C
8.
Question 8. Determine the equation of a line, in slope-intercept form, that passes through the points \((4,-9)\) and \((8,-4)\).
Solution:Find the slope:\[m=\frac{y_2-y_1}{x_2-x_1}=\frac{-4-(-9)}{8-4}=\frac{5}{4}\]Use point-slope form with \((4,-9)\):\[y-(-9)=\frac{5}{4}(x-4)\]\[y+9=\frac{5}{4}x-5\]\[y=\frac{5}{4}x-14\]Answer: D
9.
Question 9. Write an equation for a transformed sine function with an amplitude of \(\frac{5}{3}\), a period of \(\frac{1}{2}\), a phase shift of \(\frac{1}{2}\) rad to the right, and a vertical translation of 5 units down.
Solution:Use the form \(y=a\sin(b(x-c))+d\).The amplitude is \(a=\frac{5}{3}\).The period is \(\frac{1}{2}\), so:\[b=\frac{2\pi}{\text{period}}=\frac{2\pi}{\frac{1}{2}}=4\pi\]A shift \(\frac{1}{2}\) to the right gives \(x-\frac{1}{2}\).A vertical translation 5 units down gives \(d=-5\).Therefore:\[y=\frac{5}{3}\sin\left(4\pi\left(x-\frac{1}{2}\right)\right)-5\]Answer: D
10.
Question 10. Which of the following is equivalent to the expression \(\log_2(s)+4\log_2(u)+\log_2(x)\)?
Solution:Use the power rule:\[4\log_2(u)=\log_2(u^4)\]Then combine the logarithms:\[\log_2(s)+\log_2(u^4)+\log_2(x)=\log_2(su^4x)\]Answer: C
11.
Question 11. Evaluate \(\log_6(279936)\).
Solution:Rewrite \(279936\) as a power of \(6\):\[279936=6^7\]Therefore:\[\log_6(279936)=\log_6(6^7)=7\]Answer: A
12.
Question 12. Solve \(\log_3(x)=\log_3(2)+\log_3(3)\).
Solution:Use the product rule for logarithms:\[\log_3(2)+\log_3(3)=\log_3(2\cdot 3)=\log_3(6)\]So:\[\log_3(x)=\log_3(6)\]Since the bases are the same:\[x=6\]Answer: A
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