The tangent line to y= f(x) at (4,-6) passes through the point (10,9) Compute the following, f(4) and f'(4)
well you know that the slope of the tangent line at the point (4,-6) is f'(4) ?
Right!
so we can compute f'(4) using the two points the tangent line passes through : (4,-6) ,(10,9) can you do it ?
Maybe... do I just find the slope using the y-y/x-x formula and thats my answer?
yes it is ! (for f'(4))
awesome! now I'm completely lost on f(4) haha
well .. f(4) is right in front of your eyes! you got the point with x=4 that f passes through it!!
okay I'm completely confused.. I'm sorry! I'm way too new at this haha. I just do y=mx+b and plug in 4 for x, the answer to f'(4) for the slope?
f'(4) is the slope of the tangent line.. we know that the tangent line passes through the points : (4,-6) ,(10,9) so .. all you have to do is calculate the slope of the line that passes through those two points and get f'(4) (just as you said up there)
f'(4) = (9 - (-6)) / (10 - 4) = 15 / 6 = 2.5
you DONT need to find the line equation y=mx+b for this question.. you just need the slope of the tangent line (f'(4))
Alright, I think I get it! Thanks a bunch!
wait.. do you know what is f(4) ?
it's easier than f'(4)
it's given in the question!
-6?
Thank you :)
yes.. very good :)
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