Name
January 6, 2017
Honors Math 2 application and review problems
Quadratic Functions (3.01–3.08) page 1
Quadratic application and review problems
1. The Gateway Arch is the centerpiece of the St. Louis skyline. Its shape is a parabola.
Suppose the quadratic function describing this shape is
F(x) = − 701 x 2 + 6 x .
a. Find the zeros of F(x).
b. Use the zeros to find the coordinates of the vertex.
c. Sketch a graph of F(x). Label the zeros and the vertex with their
coordinates.
d. What is an appropriate choice of domain for F(x) to describe the
Arch?
e. Check your work so far by graphing F(x) = − 701 x 2 + 6 x on your calculator (you will
need to set an appropriate window), then finding the maximum on your calculator.
f. Visitors to the Arch can take a tram ride up to the top of the monument. How high above
the ground is this?
2. You are selling cookies at a school club bake sale. Naturally, the higher you set the selling
price, the fewer cookies you will sell. Suppose that if the selling price is set at x cents, you
will sell (500 – 4x) cookies. Let f(x) stand for the total amount made by the sale.
a. Write a function formula for f(x).
Parts b through e have you find the coordinates of the vertex four different ways.
b. Find the zeros of f(x), then use the zeros to find the coordinates of the vertex.
c. Rewrite the formula for f(x) in the ax2 + bx + c form, then use the −2 ab formula to find the
coordinates of the vertex.
d. Rewrite the formula for f(x) in vertex form, and use this to identify the vertex.
e. On a graphing calculator, graph f(x) and find its vertex.
f. You should have gotten the same x and y coordinates for the vertex from the first four
parts of this problem. Do all your answers agree?
g. What do the coordinates of the vertex tell you about the cookie sale?
3. a. Use completing the square to solve –2x2 + 4x + 6 = 0.
b. Using completing the square, put y = –2x2 + 4x + 6 into vertex form and find the vertex.
c. Compare the calculations used for parts a and b. What steps were the same? What steps
were different?
4. Suppose that (x – k) is one of the factors of a quadratic function f(x). What else does this tell
you about f(x) and its graph? What can you not discern from this information? Be specific.
5. Give an example of a quadratic whose zeros fit each description, or explain why you can’t.
a. only one real zero
b. two real zeros
6. Find a quadratic function whose zeros would be
c. three real zeros
1
2
and –3, and has a y-intercept of 6.
Name
January 6, 2017
Honors Math 2 application and review problems
Quadratic Functions (3.01–3.08) page 2
7. Solve for x by completing the square: 0 = 2x 2 + kx − m
8. For a quadratic function y = ax2 + bx + c, under what circumstances would c represent both
the y-intercept and the y-coordinate of the vertex?
€ that rises more steeply than y = x2 and has its vertex at
9. Write an equation for a parabola
(2, 5).
10. Solve these equations. Practice using a variety of methods.
a. 2x 2 + 7x + 3 = 0
b. 5x 2 + 17x = −6
c. 4 x 2 + 9 = 12x
d. 7x 2 −18x + 11 = 0
e. 9x 2 + 30x + 25 = 0
f. 4 x 2 − 6x + 49 = 0
2
€ 11. Suppose the equation x −15x
€ + 40 = 0 has roots α and€β. Without solving the equation,
find a quadratic equation with the following roots: −2α and −2 β.
€
€
€
Answers
€
€
€
1. a. x = 0, x = 420
b. (210, 630) €
d. 0 ≤ x ≤ 420
€
e. window 0 ≤ x ≤ 420 0 ≤ y ≤ 630
f. 630 feet
2. a. f(x) = x(500–4x)
b-f. (62.5, 15625)
g. The club could earn the most by selling cookies at 62.5 cents each, making $156.25.
3. a. −2x 2 + 4 x + 6 = 0
b.
y = −2x 2 + 4 x + 6
Vertex (1,8)
x = 3, x = −1
4. If (x − k) is a factor, then x = k is an x-intercept of the graph (and a solution, zero, and root).
€ or if there is one other factor. You also don’t
You cannot discern if this is the only factor
know the leading coefficient, a.
€
€
2
b. For example, y = (x − 2)(x − 3) c. impossible
€ 5. a. For example,€y = (x − 2)
€
1
1
3
6. y = a(x − 2 )(x + 3) → 6 = a(0 − 2 )(0 + 3) → 6 = − 2 a → a = −4
Answer: y = −4(x − 12 )(x + 3) or y = −4 x 2 −10x + 6
€
€
2
2
k
m
k2
m
k2
7. 2x + kx − m = 0 → x + 2 x = 2 → x 2 + k2 x + 16
=
+
€
2
16
€
€
€
x + 4k = ±
€
k 2 +8m
4
8. When b=0
€
€
10. a. x = − 12 ,−3
€
11
x
=
,1
7
d.
€
2
(x + 4k )
2
→ x = −k ± k4 +8m .
€
9. For example, y = 2(x − 2) 2 + 5
€
€
b. x = − 25 ,−3
c. x =
− 53
e. x = €
f. x =
2
€ 11. x + 30x + 160 = 0 €
€
€
€
€
3
2
3
4
=
k 2 +8m
16
→