Posted by Steff on .
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 reworked 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,
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,
Thanks. I was missing the step where you plug in the 1 to get slope.