College Calculus
posted by NYCalc on .
Let f(x) = 3x^2  2x + 1
a) write an equation of the line going through the points (1, f(1)), and (2, f(2)).
b) Find a point on the graph of f where the tangent line to the graph has the same slope as the line in part (a). Write the equation of the tangent line at that point.
I got the answer to part A. For part b, I'm confused...isn't the slope of the tangent line 1/slope of normal line?
for part a) i got y = 7x 5

which means you evaluated the slope as 7. So where else is it seven?
f'=6x2=7
x=1.5
y=3(9/6)^22(9/6)+1 figure that out.
Now , you know x,y and the slope, so figure the intercept b.
y=mx+b 
Thank you!