1.
Question 1. Find the zeroes of the following equation: \(y=x^2+3x-10\).
Solution: Factor the quadratic equation. \[y=x^2+3x-10\] \[0=x^2+3x-10\] \[0=(x+5)(x-2)\] So the zeroes are \(x=-5\) and \(x=2\). Answer: D
2.
Question 2. The domain of \(y=\frac{3}{(x+2)^2}-4\) is:
Solution: The denominator cannot be zero. \[(x+2)^2\neq 0\] \[x+2\neq 0\] \[x\neq -2\] Therefore, the domain is all real numbers except \(-2\). Answer: A
3.
Question 3. For the equation \(y=x^3-4x\), determine the roots (or zeros) of the equation.
Solution: Set the function equal to zero and factor. \[0=x^3-4x\] \[0=x(x^2-4)\] \[0=x(x-2)(x+2)\] Thus, \(x=0\), \(x=2\), and \(x=-2\). Answer: E
4.
Question 4. The number of solutions of \(x^2=5\cos(3x+1)\) in the interval \([-2\pi,2\pi]\) is: (Hint: You will need your graphing calculator for this one.)
Solution: Move all terms to one side and define a function. \[x^2=5\cos(3x+1)\] \[0=5\cos(3x+1)-x^2\] Let \(f(x)=5\cos(3x+1)-x^2\). Using a graphing calculator on \([-2\pi,2\pi]\), the graph crosses the x-axis 4 times. Answer: D
5.
Question 5. The graph of \(y=(x-3)^7\) is reflected over the x-axis. Which of the following is the new equation?
Solution: A reflection over the x-axis multiplies all output values by \(-1\). Original function: \(y=(x-3)^7\). Reflected function: \(y=-(x-3)^7\). Answer: A
6.
Question 6. If \(f(x)=3x^2-8x\), then \(f(2a)=12a^2-16a\).
Solution: Substitute \(2a\) into \(f(x)=3x^2-8x\). \[f(2a)=3(2a)^2-8(2a)\] \[f(2a)=12a^2-16a\] The statement is true. Answer: A
7.
Question 7. If \(f(x)=3x^3+1\) and \(g(x)=2x-1\), then \(g(f(-1))=-5\).
Solution: First find \(f(-1)\). \[f(-1)=3(-1)^3+1=-3+1=-2\] Then substitute into \(g(x)\). \[g(f(-1))=g(-2)=2(-2)-1=-5\] The statement is true. Answer: B
8.
Question 8. Which of the graphs shown below represents the base function \(f(x)=x^2\) and the stretched function \(g(x)=-\frac{1}{5}x^2\)?
Solution: The negative sign reflects the parabola over the x-axis. Since \(|a|=\frac{1}{5}<1\), the graph is vertically compressed, making it wider. Graph A shows a downward-opening wider parabola compared with \(f(x)=x^2\). Answer: A
9.
Question 9. Given the functions \(f(x)=x^2-6\) and \(g(x)=2-x\), determine an equation for the combined function \(h(x)=f(g(x))\).
Solution: Substitute \(g(x)=2-x\) into \(f(x)=x^2-6\). \[h(x)=f(g(x))\] \[h(x)=f(2-x)\] \[h(x)=(2-x)^2-6\] \[h(x)=4-4x+x^2-6\] \[h(x)=x^2-4x-2\] Answer: C
10.
Question 10. Given that \(f(x)=-6x+2\) and \(g(x)=9x^2+4x\), find \((f\circ g)(7)\).
Solution: First find \(g(7)\). \[g(7)=9(7)^2+4(7)=9(49)+28=441+28=469\] Now substitute into \(f(x)\). \[(f\circ g)(7)=f(469)\] \[f(469)=-6(469)+2=-2814+2=-2812\] Answer: D
11.
Question 11. Determine the range of the linear relation graphed below.
Solution: From the graph, the highest point is a closed point at \(y=2\). The line then continues downward forever. Therefore, the range includes 2 and all y-values less than 2. \[y\le 2\] Answer: C
12.
Question 12. Determine the domain of the relation graphed below.
Solution: From the graph, the x-values start at \(x=-4\) with an open circle, so \(-4\) is not included. The graph ends at \(x=2\) with a closed circle, so \(2\) is included. Therefore, the domain is \((-4,2]\). Answer: C
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Question 13. A bottle is riding the waves at a beach. The bottle's up and down motion with the waves can be described using the formula \(h=3.2\sin\left(\frac{\pi t}{3}\right)\), where \(h\) is the height, in metres, above the flat-water surface and \(t\) is the time, in seconds. When is the first time, to the nearest tenth of a second, that the height of the bottle will be \(2.9\) m? Hint: If you do not remember how to solve these algebraically, use a graphing calculator. Either way, be sure to be in Radian Mode.
Solution: Substitute \(h=2.9\) into the equation and solve for the first positive time. \[2.9=3.2\sin\left(\frac{\pi t}{3}\right)\] \[\frac{2.9}{3.2}=\sin\left(\frac{\pi t}{3}\right)\] \[\sin^{-1}\left(\frac{2.9}{3.2}\right)=\frac{\pi t}{3}\] \[t=\frac{3}{\pi}\sin^{-1}\left(\frac{2.9}{3.2}\right)\approx 1.1\] The first time is about \(1.1\) seconds. Answer: C
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