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

Find the derivative of y (you should know how to do this):
y' = 2x - 2
Find the slope at 2 (derivative of the function gives you the slope at that point):
y' (2) = 2(2) - 2
y' (2) = 2

Now find the equation of the line which is tangent at P:
Remember y = mx + b?
x = 2 (given)
y = -3 (given)
m = 2 (m is the slope we calculated)

Now plug it into the equation to find b:
-3 = 2(2) + b
-7 = b
Now that you know the values, form the equation:
y = 2x - 7

That's it!

Answer 2

If you learned calculus, then this answer would be resolved with the first derivative, that is, y'=2x-2. Thus at the point x=2, y' = 2. Thus your answer for the slope of the tangent is right, since y' = 2*2-2. Now that you know the slope of your tangent line is 2, you have y = 2x + b as its equation, thus since this line goes through the point P(2,-3), -3 = 2(2) + b which means that b = -7. All in all, you have the equation of the tangent line being y = 2x-7.

Answer 3

The slope of the tangent to a curve at a given point is the value of the derivative at that`point
y=x^2-2x-3

y´= 2x-2

At point(2,-3) of the curve y´=2*2-2=2

so the slope is 2 and the equation is y+3=2*(x-2)which comes to y= 2x-7

Answer 4

i am no math wiz and havent taken math in a while but i think this might be it or at least this is what i got:

y=x^2-2x+3 P(2,-3)
therefore:
-3=2^2-2(2)+3
-3=4-4+2
-3=2 -3=y y=2