1.
Question 1. What are the domain and range of the function shown in the graph?
Step 1: The graph is a parabola opening upward. Step 2: A vertical parabola continues forever left and right, so the domain is all real numbers. Step 3: The lowest point is the vertex with \(y=-4\). Step 4: Since the graph opens upward, the range is \(y\geq -4\). Answer: D: all real numbers; R: \(y\geq -4\)
2.
Question 2. What is the equation of the axis of symmetry for the parabola \(y=\frac{1}{3}(x-5)^2-2\)?
Step 1: The vertex form is \(y=a(x-p)^2+q\). Step 2: The axis of symmetry is \(x=p\). Step 3: In \(y=\frac{1}{3}(x-5)^2-2\), \(p=5\). Answer: \(x=5\)
3.
Question 3. The parabola \(y=x^2\) is changed to the form \(y=a(x-p)^2+q\) by translating the parabola 4 units down and 2 units right and expanding it vertically by a factor of 3. What are the values of \(a\), \(p\), and \(q\)?
Step 1: A vertical expansion by a factor of \(3\) gives \(a=3\). Step 2: A translation 2 units right gives \((x-2)^2\), so \(p=2\). Step 3: A translation 4 units down gives \(q=-4\). Answer: \(a=3, p=2, q=-4\)
4.
Question 4. Which of the following represents the graph of \(y=\frac{1}{2}(x-4)^2+2\)?
Step 1: The function is in vertex form \(y=a(x-p)^2+q\). Step 2: Here \(a=\frac{1}{2}\), \(p=4\), and \(q=2\). Step 3: The vertex is \((4,2)\). Step 4: Since \(a>0\), the parabola opens upward. Step 5: Since \(0<a<1\), it is wider than \(y=x^2\). Answer: Graph D
5.
Question 5. Write the equation of the parabola that opens up, has a vertex at \((3,-3)\), and is congruent to \(y=\frac{1}{3}x^2\). Answer in the form \(y=a(x-h)^2+k\).
Step 1: Vertex form is \(y=a(x-h)^2+k\). Step 2: The vertex \((3,-3)\) gives \(h=3\) and \(k=-3\). Step 3: Congruent to \(y=\frac{1}{3}x^2\) means \(a=\frac{1}{3}\). Answer: \(y=\frac{1}{3}(x-3)^2-3\)
6.
Question 6. What is the equation of the parabola with vertex \((1,3)\) and passing through \((3,5)\)?
Step 1: Use vertex form \(y=a(x-h)^2+k\). Step 2: The vertex \((1,3)\) gives \(y=a(x-1)^2+3\). Step 3: Substitute \((3,5)\): \(5=a(3-1)^2+3\). Step 4: \(5=4a+3\). Step 5: \(a=\frac{1}{2}\). Answer: \(y=\frac{1}{2}(x-1)^2+3\)
7.
Question 7. What is the vertex form of the quadratic function \(y=x^2-7x-20\)?
Step 1: Start with \(y=x^2-7x-20\). Step 2: Half of \(-7\) is \(-\frac{7}{2}\), and its square is \(\frac{49}{4}\). Step 3: Add and subtract \(\frac{49}{4}\): \(y=x^2-7x+\frac{49}{4}-\frac{49}{4}-20\). Step 4: Rewrite: \(y=\left(x-\frac{7}{2}\right)^2-\frac{49}{4}-\frac{80}{4}\). Step 5: Combine constants: \(-\frac{129}{4}\). Answer: \(y=\left(x-\frac{7}{2}\right)^2-\frac{129}{4}\)
8.
Question 8. What does the quadratic equation \(y=4x^2+24x-11\) look like when it is rewritten in the form \(y=a(x-p)^2+q\)?
Step 1: Factor \(4\) from the quadratic and linear terms: \(y=4(x^2+6x)-11\). Step 2: Complete the square: \(x^2+6x=(x+3)^2-9\). Step 3: Substitute: \(y=4((x+3)^2-9)-11\). Step 4: Simplify: \(y=4(x+3)^2-36-11\). Step 5: \(y=4(x+3)^2-47\). Answer: \(y=4(x+3)^2-47\)
9.
Question 9. Use completing the square to rewrite \(y=x^2+4x-13\) in vertex form. Identify the maximum or minimum value.
Step 1: Start with \(y=x^2+4x-13\). Step 2: Complete the square: \(x^2+4x=(x+2)^2-4\). Step 3: Substitute: \(y=(x+2)^2-4-13\). Step 4: Simplify: \(y=(x+2)^2-17\). Step 5: The parabola opens upward, so the vertex gives a minimum value. Answer: Minimum value of \(-17\).
10.
Question 10. The profit, \(P\) dollars, earned by a company producing CDs is approximated by \(P=-30x^2+720x-1720\), where \(x\) dollars is the selling price of one CD. What selling price gives the maximum profit? Solve algebraically, not graphically.
Step 1: For \(P=ax^2+bx+c\), the maximum occurs at \(x=-\frac{b}{2a}\). Step 2: Here \(a=-30\) and \(b=720\). Step 3: Substitute: \(x=-\frac{720}{2(-30)}\). Step 4: Simplify: \(x=12\). Step 5: Since \(a<0\), this gives the maximum profit. Answer: \($12\)
11.
Question 11. A farmer wants to put a fence around a vegetable garden. Only three sides must be fenced, since a rock wall will form the fourth side. If he uses \(40\text{ m}\) of fencing, what is the maximum area of the enclosed vegetable garden?
Step 1: Let the two equal widths be \(w\), and let the length be \(L\). Step 2: Since only three sides are fenced, \(2w+L=40\). Step 3: Solve for \(L\): \(L=40-2w\). Step 4: Area is \(A=wL=w(40-2w)=-2w^2+40w\). Step 5: The maximum occurs at \(w=-\frac{40}{2(-2)}=10\). Step 6: Then \(L=40-2(10)=20\). Step 7: Maximum area is \(10\cdot20=200\). Answer: \(200\text{ m}^2\)
12.
Question 12. The Brewster charter bus company provides a bus trip for a fare of \($80\) each for \(30\) or fewer passengers. For each passenger in excess of \(30\), the fare is decreased \($2\) per person for everyone. What number of passengers will produce the maximum revenue for the bus company?
Step 1: Let \(x\) be the number of passengers above \(30\). Step 2: Number of passengers is \(30+x\). Step 3: Fare per passenger is \(80-2x\). Step 4: Revenue is \(R=(30+x)(80-2x)\). Step 5: Expand: \(R=-2x^2+20x+2400\). Step 6: The maximum occurs at \(x=-\frac{20}{2(-2)}=5\). Step 7: Total passengers are \(30+5=35\). Answer: \(35\)
1 out of 1
