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Answer 1 · 2017-11-11T17:28:28

$(-3, -16)$

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 · 2017-11-11T17:32:42

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

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.

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