Question

Rewrite the quadratic function from standard form to vertex form.

f(x)x^2+6x-5

Answer 1

`y = {(x + 3)}^(2) - 14`

Explanation:

The original equation

`f(x) = {x }^(2) + 6x - 5`

vertex form

`y = a {(x - h)}^(2) + k \\ where \: (h. \: k) \: is \: the \: vertex`

solve for the x coordinate of the vertex, h

`x = h = - (b)/(2a) = - (6)/(2(1)) = - (6)/(2) = - 3 \\ x = h = - 3`

substitute -3 back into the original equation to solve for y

`y = { (- 3)}^(2) + 6( - 3) - 5 \\ y = 9 - 18 - 5 \\ y = 4 - 18 \\ y = k = - 14`

substitute h and k values into the vertex form

a = 1 So we can omit it

`y = {(x - ( - 3)}^(2) - 14 \\ y = {(x + 3)}^(2) - 14`