# Thread: Need help with tangent lines and derivatives

1. ## Need help with tangent lines and derivatives

Hey,

I'm struggling with this one problem from my text book.
"Find the equations of the tangent lines to the parabola y=(x^2)+x that pass through the point (2, -3)."

I know how to find the derivative (2x+1) and then I can find one set of points (5,30) but I can't figure out how to find the other set of points.

2. ## Re: Need help with tangent lines and derivatives

f'(a) = 2a + 1

And the slope of the line between the point (a, a^2 + a) and (2, -3) is (a^2 + a + 3)/(2 - a) = 2a + 1
Can you solve that?

3. ## Re: Need help with tangent lines and derivatives

Yes, I get the points (5, 30) and (-1,0) and from there I can find the equations of the lines. Thank you so much!

4. ## Re: Need help with tangent lines and derivatives

Originally Posted by newslang
Hey,

I'm struggling with this one problem from my text book.
"Find the equations of the tangent lines to the parabola y=(x^2)+x that pass through the point (2, -3)."

I know how to find the derivative (2x+1) and then I can find one set of points (5,30) but I can't figure out how to find the other set of points.

Let's say you have a point (p,r) on your parabola. (So, r is given by $r=p^2+p\,.$ )

The slope, m, of the line tangent to the parabola when x = p is $m=2p+1$. (this from your derivative)

The slope, m, of the line passing through points (p, r) and (2, -3) is given by $m=\frac{r-\,-3}{p-2}=\frac{r+3}{p-2}\,.$

So that $m(p-2)=r+3\,.$

Plug the in the earlier results for r & m into this. Solve for p. (There will likely be two answers.) You then have the x value(s) for the intersection of the tanget line, with the parabola.

5. ## Re: Need help with tangent lines and derivatives

Originally Posted by newslang
Hey,

I'm struggling with this one problem from my text book.
"Find the equations of the tangent lines to the parabola y=(x^2)+x that pass through the point (2, -3)."

I know how to find the derivative (2x+1) and then I can find one set of points (5,30) but I can't figure out how to find the other set of points.

Another way to do a problem like this is to use Fermat's "method of ad-equations" which predates Calculus. Any line through (2, -3) can be written y= m(x-2)- 3= mx- 2m- 3. Of course, at a point where such a line crosses the graph of $y= x^2+ x$ the two functions must give the same y-value: $mx- 2m- 3= x^2+ x$ or $x^2+ (1- m)x+ 2m+ 3= 0$. That's a quadratic equation with solution given by the quadratic formula: $x= \frac{m-1\pm\sqrt{(1- m)^2- 4(2m+ 3)}}{2}$. But if it is tangent that x must be a double root- the only root of this equation. That means that the discriminant must be 0: $(1- m)^2- 4(2m+3)= 0$. Solve that equation for m and then $x= \frac{m-1}{2}$.