Question

(Calculus) Find the slope of the curve at the given point p and an equation of the tangent line at p. y=(x^2)+3, P (2,1). I've managed to [somehow] find the slope of the secant line, which is 4+h. I cannot figure how the book is coming up with the tangent line of y=4x-7.

Answer 1

Because you'll notice that when x=2, y=7. As stated by P(2,1)
Therefore, do the basic algebra on that, you'll see where y=4x - 7.

Answer 2

No I don't notice that. Because P is (2,1), x=2 means y=1. Or do you mean \[x _{2}=2\] Then I'm guessing I have to do the tables? Or... what algebra did you use to get the 7?

Answer 3

remember that, by the fundemental thereom of calculus, your taking the limit as h approaches 0.

Answer 4

oh yeah, Prof pointed that out in lecture this morning. She just didn't go over anything on how to write an equation for the tangent of the parabola at point P. That's where I'm getting lost?

Answer 5

If the point is P=(2,1). The point is outside the parabola and there will be two tangents from this point to the curve. The equations are
\[
y_1= -2 \left(\sqrt{6}-2\right) x+4
\sqrt{6}-7\\
y_2=2 \left(2+\sqrt{6}\right) x-4
\sqrt{6}-7
\]

Answer 6

See the graph

Answer 7

The book states there is only one tangent line. I'm also wondering where you got the square root of 6?

Answer 8

There might be a misprint in your problem.

Answer 9

There is a way to find the two tangents not requiring calculus. I think you worded your problem wrongly. If a point is outside the parabola, there are two tangents from this line to to the parabola. See the graph above.

Answer 10

I am going to a meeting. See you later.

Answer 11

Oh! I just got it. I was missing the equation of \[y-y _{1}=m(x-x _{1})\] for the equation of the tangent at given point P. Plugging in the slope of the curve 4 and the first given points, I arrived at the correct equation.
I don't believe I worded the problem wrong, as I wrote it out just as the book had it.
"In the exercises find (a) the slope of the curve at the given point P and (b) an equation of the tangent line at P." I was then given \[y=x ^{2}-3, P (2,1)\]
I'm sorry the original wasn't as clear, the set-up of this website is still new and I couldn't find how to do the Equation input on the original question.

Answer 12

Your original parabola was
\[
y=x^2 +3
\]
So you did not type the problem correctly.