Question

Write the equation of the tangent line to the graph of f at the given point / x-value
f(x)=x^2+2x+1 @ (-3,4)
Need someone to do the process because I don't understand how to solve it.

Answer 1

Would you please find the derivative of your function f(x)?
That derivative represents the slope of the tangent line to the curve.

Answer 2

f'(x)=2x+2

Answer 3

Cool. So, f'(x) = 2x + 2.
Now please let x = -3. (Why?)

Answer 4

umm to find the slope....

Answer 5

Yes. But why use x=-3?

Answer 6

i'm not sure

Answer 7

I wanted you to make that connection. You're given the point (-3,4) and want to find the equation of the tangent line to the graph at that point.
That's where the -3 comes from. You're looking to write the equation of the tangent line to the graph at (-3,4), right?

Answer 8

yes

Answer 9

Cool. so, if the derivative f'(x) is 2x+2, and x=-3, evaluate f'(-3).

Answer 10

f'(-3)=-4

Answer 11

That's it. What's the significance of this result?

Answer 12

I suppose that -4 is the slope therefore if we put it in slope format y-y=m(x-x) the answer will be y-4=-4(x+3)

Answer 13

That's right! -4 is the slope. Perfect. Please simplify your result. Good going!!

Answer 14

Thank you

Answer 15

Simplify y-4=-4(x+3): y=4-4(x+3). Care to take the simplification one step further?

Answer 16

no this is fine, thank you