Posted by **Steff** on Wednesday, June 15, 2011 at 7:54am.

Find the slope and an equation of the tangent line to the graph of the function f at the specified point.

f(x)=-1/3x^2+5x+5: (-1, -1/3)

Answer: f’(x)=-2/3x + 5

y=-2/3x -1

I re-worked the problem and got f'(x)=5, y=5x+ 14/3

This is from a multiple choice test and this is not an answer.

Can someone check my work?

- Calculus derivative -
**Reiny**, Wednesday, June 15, 2011 at 8:48am
Your derivative is correct

Now use the value of x of the given point to find the slope

slope = (-2/3)(-1) + 5

= 2/3 + 5 = 17/3

equation is

y = (17/3)x + b

sub in the point

-1/3 = (17/3)(-1) + b

16/3 = b

equation: y = (17/3)x + 16/3

- Calculus derivative -
**Steff**, Wednesday, June 15, 2011 at 8:59am
Thanks. I was missing the step where you plug in the -1 to get slope.

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