Find an equation for the tangent to the curve at the given point.

y=x^2-x,(4,12)

Any ideas on how to solve this problem?

Question

Find an equation for the tangent to the curve at the given point.

y=x^2-x,(4,12)

Any ideas on how to solve this problem?

anonymous · Answer

take the derivative first

anonymous · Answer

2x?

anonymous · Answer

y'=2x-1

anonymous · Answer

so at that point y'(4) will give you the slope of the line

abb0t · Answer

@Brandon77  is correct. Find the slope and then plug in the given point for x. That's your slope (m) at that given point. Now that you have your slope. Put it into the slope equation.

anonymous · Answer

Ah, okay. Then I use y = mx +b and find b.

anonymous · Answer

you got it

abb0t · Answer

I would leave it as is unless they ask you to re-write it in standard form.. I mean, i'm assuming this is a Calculus course and it's assumed that you've mastered your algebra.

anonymous · Answer

You can also put it into $$y-y_{1}=m(x-x _{1}) $$ and you wouldn't haven't to find b.

anonymous · Answer

Okay. Either way the final answer is y=7x-16.

anonymous · Answer

You are correct.