1. ## Finding tangent equations

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

I found the derivative which is 2x + 1 then I'm stuck.
Any help/advice are greatly appreciated ! Thanks !

2. Originally Posted by letzdiscuss
Finding tangent equations of both lines through the point (2,-3) that are tangent to the parabola y=x^2 + x.

I found the derivative which is 2x + 1 then I'm stuck.
Any help/advice are greatly appreciated ! Thanks !

Equation of a line in point slope form

$
y-y_0=m(x-x_0)
$

You are given a point and the other point is $(x,x^2+x)$

$

x^2+x-(-3)=(2x+1)(x-2)
$
is what you get when you substitute all the given info into the point slope form

Solving that will give you what x values make a tangent line that go through your point

3. Originally Posted by letzdiscuss
Finding tangent equations of both lines through the point (2,-3) that are tangent to the parabola y=x^2 + x.

I found the derivative which is 2x + 1 then I'm stuck.
Any help/advice are greatly appreciated ! Thanks !
The equation of a line tangent to the parabola at (a, a^2 + a) is given by y = (2a + 1)(x - a) + a^2 + a = (2a + 1)x - a^2. This line goes through (2, -3) iff (2a + 1)*2 - a^2 = 3. So you need to solve the quadratic equation a^2 - 4a + 1 = 0 for a and plug the answers into the equation for the line.

4. Originally Posted by rn443
The equation of a line tangent to the parabola at (a, a^2 + a) is given by y = (2a + 1)(x - a) + a^2 + a = (2a + 1)x - a^2. This line goes through (2, -3) iff (2a + 1)*2 - a^2 = 3. So you need to solve the quadratic equation a^2 - 4a + 1 = 0 for a and plug the answers into the equation for the line.
You mean $(2a + 1)*2 - a^2 = -3$

5. Originally Posted by artvandalay11
You mean $(2a + 1)*2 - a^2 = -3$
Ah, whoops, yes.