Question

The function f(x) = x^2 +8x+10 is equivalent to the function f(x)=(x-h)^2+k. What is the value of h and k . At what value of x is the minimum value of the function

Answer 1

ANSWER

$h = - 4,k = - 6$

Minimum value: y=-6

Occurs at: x=-4

EXPLANATION

Given

$f(x) = {x}^{2} + 8x + 10$

We need to write the equivalent of this function in vertex form:

$f(x) = ({x - h)}^{2} + k$

where (h,k) is the vertex.

We must complete the square to get the function to this form.

We add and subtract the square of half the coefficient of x.

$f(x) = {x}^{2} + 8x +( \frac{8}{2}) ^{2} - ( \frac{8}{2}) ^{2} + 10$

This gives us;

$f(x) = {x}^{2} + 8x +16- 16 + 10$

The first three terms form a perfect square quadratic trinomial.

$f(x) =( x + 4) ^{2} - 6$

or

$f(x) =( x - - 4) ^{2} - 6$

Therefore we compare to

$f(x) = ({x - h)}^{2} + k$

h=-4 and k=-6

The minimum value of the function is

$y = - 6$

and this occurs at;

$x = - 4$