How do you find the slope of the tangent line to the graph of the function h(t)=t^2+3h(t)=t2+3 at (-2,7)?

1 Answer
Nallasivam V
Jun 30, 2017

The equation of the tangent is -

y=-4x-1y=−4x−1

Explanation:

Given -

h(t)=t^2+3h(t)=t2+3

Let us have it as -

y=t^2+3y=t2+3

Its slope at any point is given by its first derivative.

dy/dx=2tdydx=2t

Slope of the curve exactly at x=-2x=−2 is-

dy/dx=2(-2)=-4dydx=2(−2)=−4

The tangent is passing through the point (-2,7)(−2,7)

The slope of the tangent at Point x=-2x=−2 is the same as slope of the curve at that point (-2,7)(−2,7)

m=-4m=−4
x=-2x=−2
y=7y=7
mx+c=ymx+c=y
(-4)(-2)+c=7(−4)(−2)+c=7
8+c=78+c=7
c=7-8=-1c=7−8=−1

The equation of the tangent is -

y=-4x-1y=−4x−1

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