I am new in here and I got a problem, I am working in Derivatives: algebraic viewpoint and I have solved the equation f(x)=(x^2)-3;a=2. H Approaches 0. Now the Quotient of it is 4+h and f'(2)=lim (4+h) Well if h approaches 0 then 4 is left alone. Now here is the issue I do not know how the teacher got y-1=4(x-2) WHATS UP WITH THAT!!!

Question

anonymous · Answer

The teacher wants you to write the equation of the tangent line to the curve $$x^2-3$$ So by evaluating the limit of (4+h) as h approaches 0 you have found the slope which is 4. Now you need to find a point that the tangent line passes through so you can write the equation. A point on the tangent line can be written as (a, f(a)). So plug the value of a into the original equation like this $$(2)^2-3$$ Which equals 1. So the tangent line passes through the point (2,-1). Now use the point slope formula to write the equation of the tangent line. Point slope formula is $$y - y{1} = m(x - x{1})$$ So plug in x,y and m and you get $$y - 1 = 4(x - 2)$$ Solve for y and you get the equation of the Tangent line

anonymous · Answer

pases through the point (2,1) not -1.. sorry

anonymous · Answer

Yeah  those are the points, Thanks man then it is y=4x-7 
Dude you saved me thank you so much

anonymous · Answer

Yep that is correct.