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Lostsunrise [7]
3 years ago
14

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 i

s the minimum value of the function
Mathematics
1 answer:
topjm [15]3 years ago
3 0

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

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