Question

Which function includes the minimum or maximum value of f as a number that appears as it is shown?

a) f(x)=(x+2)^{2}−16
b) f(x)=(x−2)(x+6)
c) f(x)=x^{2}+4x−12
d) f(x)=x^{2}+6x−2x−12

Answer 1

We can find the maximum or minimum value of a quadratic function using the vertex formula:

The x-coordinate of the vertex is given by: x = -b/2a

The y-coordinate of the vertex is simply the value of the function at x = -b/2a.

So, in order for the function to include the minimum or maximum value of f as a number that appears as it is shown, we need to rewrite each of the given functions in vertex form.

a) f(x) = (x+2)^2 - 16
Vertex form: f(x) = a(x-h)^2 + k
f(x) = (x-(-2))^2 - 16
f(x) = (x+2)^2 - 16

The vertex occurs at (-2, -16), and the minimum value of the function is -16.

b) f(x) = (x-2)(x+6)
To find the vertex form, we need to expand the expression:
f(x) = x^2 + 4x - 12
Vertex form: f(x) = a(x-h)^2 + k
Completing the square: f(x) = (x+2)^2 - 16

The vertex occurs at (-2, -16), and the minimum value of the function is -16.

c) f(x) = x^2 + 4x - 12
Vertex form: f(x) = a(x-h)^2 + k
Completing the square: f(x) = (x+2)^2 - 16

The vertex occurs at (-2, -16), and the minimum value of the function is -16.

d) f(x) = x^2 + 6x - 2x - 12
Simplifying: f(x) = x^2 + 4x - 12
Vertex form: f(x) = a(x-h)^2 + k
Completing the square: f(x) = (x+2)^2 - 16

The vertex occurs at (-2, -16), and the minimum value of the function is -16.

Therefore, all the functions have the same minimum value of f as a number that appears as it is shown, which is -16.

Answer 2

The function that includes the minimum or maximum value of f as a number that appears as it is shown is option (c) f(x)=x^2+4x-12.

To find the minimum or maximum value, we can complete the square by adding and subtracting (4/2)^2 = 4 to the function to get:

f(x) = (x^2 + 4x + 4) - 4 - 12
f(x) = (x + 2)^2 - 16

Since (x + 2)^2 is always non-negative, the minimum value of f(x) is -16, which occurs when (x + 2)^2 = 0, or x = -2