Question

The vertex form of h(x) = x2 – 14x + 6 is h(x) = (x – )2 – .

Answer 1

Answer:

$h(x)=(x-7)^{2} -43$

Step-by-step explanation:

The given function is

$h(x)=x^{2}-14x+6$

The vertex form is obtained by "completing the square". First, we have to add and subtract a term, which is formed by the squared power of half the coefficient of the linear term:

$x^{2}-14x+6\\x^{2}-14x+6+(\frac{14}{2} )^{2}-(\frac{14}{2} )^{2}\\x^{2} -14x+6+7^{2}-7^{2}\\x^{2} -14x+7^{2} +6-7^{2}\\(x-7)^{2} +6-7^{2} =(x-7)^{2} +6-49\\(x-7)^{2} -43$

Therefore, the expression would be $h(x)=(x-7)^{2} -43$

Answer 2

The vertex form of h(x) = x2 – 14x + 6 is h(x) = (x –7 )2 – 43.