Question

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

Answer 1

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

Answer 2

Thank you!