Find slope of tangent line

[kickingtoad]

The question is

f(x)=-x^2-2x
Find the equation of the line tangent to the graph f(x) at the point (-1,1)

This is what I've been doing:

1. find f'(x)
f'(x)= -2x-2

2. find slope
f'(-1)=slope
f'(-1)=-2(-1)-2
f'(-1)=2-2=0
Slope is 0

3. Equation of a line is y=mx+b
How do I find the y value with the point given?
Do I fill 1 in for x?

 

[lookagain]

f(x)=-x^2-2x
Find the equation of the line tangent to the graph f(x) at the point (-1,1)

2. find slope
f'(-1)=slope
f'(-1)=-2(-1)-2
f'(-1)=2-2=0
Slope is 0

3. Equation of a line is y=mx+b
How do I find the y value with the point given?
Do I fill 1 in for x?

The y-intercept is (0, 1) because all points on that horizontal line have a y-coordinate
of 1, as it is that that tangent line passes through the given point of (-1,1). Then b = 1.

Plug in m = 0 and b = 1 into y = mx + b.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Also, note that from algebra that the vertex, \(\displaystyle ({\frac{-b}{2a}, f(\frac{-b}{2a})), \ equals \ (-1, 1).\)

It is the lowest point on the parabola, which has a horizontal slope. That is, its slope is zero.

 

[kickingtoad]

Thank you. You are correct.