I could really use some help on this slope/tangent problem, please?

Question

y = x^2 - 2x - 3
P(2,-3)

**Find the slope of the curve at the given point P, and the equation of the tangent line at P.

I pretty much know the answer is 2 and y = 2x-7 but I'd really like to know how to get there. Thanks so much.

We have not been taught how to derive yet in class so I should probably use another method.

Answer 1

the slope of the curve at a given point can be obtained by taking the derivative of curve equation wrt. x
i.e. d/dx(x^2-2x-3) = 2x-2
now put point p(2,-3) in it i.e. 2(2)-2=2
hence slope of curve at point p = 2
and equation of tangent is y-y1=m(x-x1)
where y is the tangent passing through (x1,y1) and having slope m .
thus m = 2 here and x1=2 y1=-3
putting in the equation y-(-3)=2(x-2) => y+3=2x-4=> y = 2x-7 Ans.

Answer 2

First calculate the derivative of y=x^2-2x-3:dy/dx=2x-2. When x=2, dy/dx=2--this is your slope. Next use the point slope formula:y+3=2(x-2)....y=2x-7.