Finding the slope of line tangent to a parabola

1. Sep 30, 2011

oates151

1. The problem statement, all variables and given/known data

Find the equations of both lines through the point (2,-3) that are tangent to the parabola y=(x^2)+x

2. Relevant equations

3. The attempt at a solution

Took the derivative and got a slope of 5 and the slope of the normal line being -1/5, but the answer was marked wrong. How do I do this?

Two equations I got
y=-(1/5)x-(13/5)
y=5x-13

2. Sep 30, 2011

ElijahRockers

Choose a point (a,y(a)). A line that goes through this point, AND the given point must have a slope of $\frac{y(a) - (-3)}{a - 2}$ Also, the slope at point 'a' can be given by the derivative of the function. This gives you two equal expressions for the slope in terms of a. It will be a quadratic equation. The roots will be the x values at which the lines intersect the parabola.

Last edited: Sep 30, 2011
3. Sep 30, 2011

SammyS

Staff Emeritus
What do you mean? "both lines"

One of your lines is tangent to the parabola at (2, -3) .

The other is normal to the parabola at (2, -3) .

4. Sep 30, 2011

ElijahRockers

Actually I think it's tangent to the parabola at (2, 6)

5. Sep 30, 2011

SammyS

Staff Emeritus
Ha!

Yup, the parabola doesn't pass through (2, -3) ! DUH