3.1 5, a quadratic function is given

Question

A quadratic function f is given.

f(x) = −x² + 16x

(a)

Express f in vertex form.

f(x) =  

 

(b)

Find the vertex and  x- and y-intercepts of f. (If an answer does not exist, enter DNE.)

vertex(x, y)

vertex:

X intercept (smaller x value):

X intercept (larger x value)

Y intercept:

c. Find the domain and range of x using interval notation

Answer 1

For vertex form, we need to first determine the vertex, which is the coordinates (8, 64).

1. Vertex form is a(x - h)² + k where h is our x coordinate and k is the y coordinate and a is our value in the original form (the ax²). So this in vertex form is -(x - 8)² + 64
2. Part B
3. Vertex is (8, 64): original determined it using -x²+16x so used -b/2a for the x coordinate so -16/-2 = 8. Plug in 8 to the function so y coordinate is 64.
4. smaller x intercept (0,0) larger x intercept (16,0)
5. -x² + 16x = 0
6. factor out "-x" so, -x(x-16) =
7. set both factors to zero and solve for x so, x = 0,16
8. Y intercept is (0,0) because the "c" value of the quadratic is the y intercept which in this problem is 0.
9. Domain and Range
10. Domain of any quadratic is (-∞,∞)
11. Range is (-∞,64) because the max y value is 64 ( y coordinate of vertex)

Answer 2

-x^2+ 16x

=-(x^2-16x+64)+64

= -(x-8)^2 +64

vertex= (8,64)

x intercepts = 0 and 16

y intercept = 0

domain= all real numbers

range y</= 64