Find the point on parabola y= -x^2 + 3x + 4 where the slope of the tangent line is 5.

Question

is this right? i got

y = -x^2 + 3x + 4

y' = -2x + 3 = 5

y' = -2(-1)+ 3 = 5

y' = 2 + 3 = 5

y' = 5

x = -1

y = -(1)²+3(-1)+4

y = -1 - 3 + 4

y = 0

therefore (-1, 0) is the point on the parabola

Answer 1

Indeed,the derivative is:

y'=-2x+3

This derivative is 5, when

-2x+3=5

-2x=2

x=-1

-(-1)^2+3(-1)+4=

-(1)+3(-1)+4= <-- exponents before negative!!!

-1-3+4=0

the point is(-1,0)

Nicely done, good job