Question #ff0f9

2 Answers
Nov 11, 2017

#(-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)#

Nov 11, 2017

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#