Posted by **NY-Calc** on Monday, March 7, 2011 at 12:45pm.

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

- College Calculus -
**bobpursley**, Monday, March 7, 2011 at 1:30pm
which means you evaluated the slope as 7. So where else is it seven?

f'=6x-2=7

x=1.5

y=3(9/6)^2-2(9/6)+1 figure that out.

Now , you know x,y and the slope, so figure the intercept b.

y=mx+b

- College Calculus -
**NY-Calc**, Monday, March 7, 2011 at 2:01pm
Thank you!

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