The point of the graphp of f(x)=x^2+4*x-14 at which the tangent line is perpendicular to teh line y=5*x-15 is?

Question

The point of the graphp of f(x)=x^2+4*x-14 at which the tangent line is perpendicular to teh line y=5*x-15 is (_____, _____)

Answer 1

2x+4 = -1/5
2x = -4 1/5
x = -2.1
y = -17.99

Answer 2

1) what is slope of second line, y=5*x-15 ?
It is 5, so the TL you want has a slope of -(1/5), since it's perpendicular.

2) f(x)=x^2+4*x-14 , so
f ' (x) = 2x + 4, and you want the x so that this is -(1/5), or

2x + 4 = -(1/5) or
2x = -(1/5) -4 or
2x = - 21/5 or
x= -21/10.

3) So the point on f(x) when x = -21/10 is where y =
y= f(x)=x^2+4*x-14 when x = -21/10 or

y = -(21/10)^2 + 4*(-21/10) - 14.

Clean this up, and the point is (-21/10,y).