Tangents intersect off of curve y = x^2 + x

[theschaef]

Here is the question in the text:

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

I've tried a bunch of things but have got no where.. any help would be appreciated!

 

[pka]

Suppose that \(\displaystyle \left( {a,a^2 + a} \right)\) is the point tangency on the curve.

Then the slope is \(\displaystyle \begin{array}{rcl}
\frac{{a^2 + a + 3}}{{a - 2}} & = & 2a + 1 \\
a^2 + a + 3 & = & 2a^2 - 3a - 2 \\
\end{array}.\)

Now solve for a,

 

[galactus]

Find the derivative of \(\displaystyle y=x^{2}+x\)

\(\displaystyle y'=2x+1\). There's your slope m.

Use \(\displaystyle y-y_{1}=m(x-x_{1})\)

\(\displaystyle x_{1}=2\) and \(\displaystyle y_{1}=-3\)

\(\displaystyle x^{2}+x+3=(2x+1)(x-2)\)

Solve for x.

You can then use those values and your given coordinates to find your lines. Take a stab at it and write back if you get stuck.

 

[theschaef]

Alright.. i got y=11x-25 and y=-x-1 for the tangents.. i think thats right.

 

[galactus]

Very good. Yep, that's it.

 

[theschaef]

woot.. thanks a lot!