Question

Question #ff0f9

Answer 1

`(-3, -16)`

Explanation:

first, convert `x^2+6x-7` into `p(x+q)^2+r` form

`x^2+6x-7 = (x^2+6x+9)-16`

`= (x+3)^2-16`

`y = (x+3)^2-16`
turning point (vertex) `= (-q, r)`

`q = 3`
`-q = -3`

`r = -16`

`therefore (-q ,r) = (-3, -16)`

graph{x^2+6x-7 [-5.135, -0.19, -16.167, -13.695]}

(vertex displayed at `(-3, -16)`

Answer 2

The turning point is at `(-3,-16)`

Explanation:

First derive the equation:

`x^2 +6x - 7`

`dy/dx = 2x+6`

Then, make it equal to zero since the gradient at the turning point is zero.

`2x+6 =0`

Now solve for `x`:

`2x+6=0 -> 2x=-6`

`x=-3`

To find the `y` value, substitute the `x` value into the equation of the curve.

`y = -3^2 +6(-3) -7`
`y = 9-18-7`
`y=9-25`
`y= -16`