Question

f(x) = -16x2 + 24x + 16

Part A: What are the x-intercepts of the graph of f(x)? Show your work. (2 points)

Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points)

Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph.

Answer 1

A) To get x-intercepts, set f(x) (or y) to zero and solve for x.
Solve -16x^2 + 24x + 16 = 0
divide throughout by -8
2x^2 - 3x - 2 = 0
solve for x.

Answer 2

2x^2 - 3x - 2 = 0
2x^2 - 4x + x - 2 = 0
2x(x - 2) + 1(x - 2) = 0
(2x+1)(x-2) = 0
x = -0.5 or x = 2
The x-intercepts are: -0.5 and 2.

Answer 3

B) Is the vertex of the graph of f(x) going to be a maximum or minimum?
f(x) = -16x^2 + 24x + 16
The "a" or the coefficient of x^2 is -16. Since it is a negative, it is an inverted parabola that opens downward. Therefore, the vertex will be a maximum.

To find the vertex, complete the square of -16x^2 + 24x + 16
The vertex form of this equation will look like: f(x) = a(x-h)^2 + k where (h,k) is the vertex.

Answer 4

Complete the square of -16x^2 + 24x + 16
Make the coefficient of x^2 1 by factoring out -16
-16(x^2 - 24/16x - 1) =
-16(x^2 - 3/2x - 1).
To complete the square: Divide the coefficient of x by 2: -3/2 / 2 = -3/4. This will go inside the parenthesis to be squared and you will have to subtract the square of it:
-16(x^2 - 3/2x - 1) = -16{ (x-3/4)^2 - (-3/4)^2 - 1 } =
-16{ (x-3/4)^2 - 9/16 - 1 } = -16{ (x-3/4)^2 - 25/16 } =
f(x) = -16(x-3/4)^2 + 25
compare this to f(x) = a(x-h)^2 + k where (h,k) is the vertex and you can see that h = 3/4 and k = 25.
So the vertex is (3/4, 25) or (0.75, 25)

Answer 5

C) f(x) = -16x^2 + 24x + 16
We know the graph is a parabola. We also know it is an inverted parabola because of the negative coefficient of x^2.
We know the curve crosses the x-axis at x = -0.5 and x = 2 as they are the x-intercepts found in part A).
From Part B) we know that the vertex of the parabola is at (0.75, 25).
We can also set x = 0 and find the y-intercept: -16(0)^2 + 24(0) + 16 = 16.
So the curve crosses the y-axis at y = 16
We have enough information to plot the graph: