Finding the slope of line tangent to a parabola

[oates151]

Homework Statement

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

Homework Equations

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

 

[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.

 

[SammyS]

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) .

 

[ElijahRockers]

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) .

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

 

[SammyS]

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

Ha!

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