Question

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?

Answer 1

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

Answer 2

Thanks. I was missing the step where you plug in the -1 to get slope.