1.
Question 1.
If \(f(x)=-2(x-3)^2+4\), then \(f(1)=-4\).
Solution: Step 1. Substitute \(x=1\) into the function. Step 2. \(f(1)=-2(1-3)^2+4\). Step 3. \(f(1)=-2(-2)^2+4=-2(4)+4=-8+4=-4\). Step 4. The statement is true. Answer: B
2.
Question 2.
The domain of \(y=\frac{3}{(x+2)^2}-4\) is:
Solution: Step 1. A rational function is undefined when its denominator is zero. Step 2. Set the denominator equal to zero: \((x+2)^2=0\). Step 3. \(x+2=0\), so \(x=-2\). Step 4. Therefore the domain is all real numbers except \(-2\). Answer: A
3.
Question 3.
For the equation \(y=-x^2+3\), what is the maximum value of \(y\)?
Solution: Step 1. The equation \(y=-x^2+3\) is a downward-opening parabola. Step 2. The vertex is at \((0,3)\). Step 3. Since the parabola opens downward, the maximum \(y\)-value is \(3\). Answer: A
4.
Question 4.
The number of solutions of \(x^2=5\cos(3x+1)\) in the interval \([-2\pi,2\pi]\) is:
Solution: Step 1. Rewrite the equation as \(5\cos(3x+1)-x^2=0\). Step 2. Graph \(y=5\cos(3x+1)-x^2\) on the interval \([-2\pi,2\pi]\). Step 3. Count the \(x\)-intercepts in this interval. Step 4. There are 4 solutions. 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: Step 1. A reflection over the \(x\)-axis multiplies all \(y\)-values by \(-1\). Step 2. Starting with \(y=(x-3)^7\), the reflected graph is \(y=-(x-3)^7\). Answer: B
6.
Question 6.
\(f(x)=x^2-2\) and \(g(x)=x^2-4\). Find \(g(f(x))\).
Solution: Step 1. Since \(g(x)=x^2-4\), replace \(x\) in \(g(x)\) with \(f(x)\): \(g(f(x))=(f(x))^2-4\). Step 2. Substitute \(f(x)=x^2-2\): \(g(f(x))=(x^2-2)^2-4\). Step 3. Expand: \((x^2-2)^2=x^4-4x^2+4\). Step 4. \(g(f(x))=x^4-4x^2+4-4=x^4-4x^2\). Answer: C
7.
Question 7.
If \(f(x)=3x^3+1\) and \(g(x)=2x-1\), then \(g(f(-1))=-5\).
Solution: Step 1. First find \(f(-1)\). Step 2. \(f(-1)=3(-1)^3+1=-3+1=-2\). Step 3. Now substitute into \(g(x)\): \(g(f(-1))=g(-2)=2(-2)-1=-4-1=-5\). Step 4. The statement is true. Answer: A
8.
Question 8. Which of the graphs shown below represents the base function \(f(x)=x^2\) and the stretched function \(g(x)=-\frac{7}{2}x^2\)?
Solution: Step 1. The negative sign means the graph is reflected over the \(x\)-axis. Step 2. Since \(\left|-rac{7}{2} ight|>1\), the graph is vertically stretched, making it narrower. Step 3. The graph that is reflected downward and narrower is option D. Answer: D
9.
Question 9.
Given \(f(x)=x^2-4\) and \(g(x)=7-x\), determine an equation for the combined function \(h(x)=f(g(x))\).
Solution: Step 1. \(h(x)=f(g(x))=(g(x))^2-4\). Step 2. Substitute \(g(x)=7-x\): \(h(x)=(7-x)^2-4\). Step 3. Expand: \((7-x)^2=x^2-14x+49\). Step 4. Therefore \(h(x)=x^2-14x+49-4=x^2-14x+45\). Answer: A
10.
Question 10.
Given that \(f(x)=5x+6\) and \(g(x)=-5x^2+6x\), find \((f\circ g)(-7)\).
Solution: Step 1. \((f\circ g)(-7)=f(g(-7))\). Step 2. \(g(-7)=-5(-7)^2+6(-7)=-5(49)-42=-245-42=-287\). Step 3. \(f(-287)=5(-287)+6=-1435+6=-1429\). Answer: A
11.
Question 11. Determine the range of the linear relation graphed below.
Solution: Step 1. From the graph, the highest \(y\)-value is \(2\). Step 2. The point at \(y=2\) is closed, so \(2\) is included. Step 3. The graph continues downward, so all \(y\)-values less than \(2\) are included. Step 4. Therefore the range is \(y\le 2\). Answer: C
12.
Question 12. Determine the domain of the relation graphed below.
Solution: Step 1. From the graph, the \(x\)-values go from \(-4\) to \(2\). Step 2. The point at \(x=-4\) is open, so \(-4\) is not included. Step 3. The point at \(x=2\) is closed, so \(2\) is included. Step 4. Therefore the domain is \((-4,2]\). Answer: D
13.
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 \(h=-3.6\sin\left(\frac{\pi t}{3}\right)\), 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 \(-1.3\text{ m}\)? Hint: If you don’t remember how to solve these algebraically, use a graphing calculator! (Either way, be sure to be in Radian Mode!)
Solution: Step 1. Set \(h=-1.3\): \(-1.3=-3.6\sin\left(rac{\pi t}{3} ight)\). Step 2. Divide by \(-3.6\): \(\sin\left(rac{\pi t}{3} ight)=rac{1.3}{3.6}\). Step 3. Take inverse sine: \(rac{\pi t}{3}=\sin^{-1}\left(rac{1.3}{3.6} ight)\). Step 4. Solve for \(t\): \(t=rac{3}{\pi}\sin^{-1}\left(rac{1.3}{3.6} ight)pprox 0.4\). Answer: A
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