Question

Find the equation of the line tangent to the given curve with the given slope

y=(1/3)x^3+2x^2+5, m=-4

I'm stuck. So far, I have:

y'=x^2+4x

x^2+4x=-4

Answer 1

y' = x^2 + 4 x = slope of curve = -4 at our spot
so
x^2 + 4 x + 4 = 0
(x+2)^2 = 0
x = -2
so when x = -2
y = -8/3 + 8 + 5 = (-8 +39)/3 = 31/3

so line of slope = -4 through (-2 , 31/3)
y = -4 x + b
31/3 = -4(-2) + b
b = 31/3 - 24/3 = 7/3
y = -4 x + 7/3
3 y = -12 x + 7

Answer 2

the line tangent is y=mx+b
you know m=-4
y=-4x+b is the line.

Now, we know dy/dx=4 at the tangent point, or
4=x^2+4x
(x^2+4x+4)=0 or (x+2)^2=0 or x=-2
and y=(1/3)x^3+2x^2+5= (1/3)(-2)^3+2(-2)^2 +5=10.33.
so the point of tangency is -2,10.33
y=mx+b
10.33=-4(-2)+b solve for b, and you have the equation of the line.