Tangent Lines Crossing through a Specified Point

[jpnov]

For this one, I know that you must start out by taking the derivative of y.

y' = 2x

I then plugged the x value in, to get an m value, but I don't think that's right, because I'm not finding the slope at (2, -3), I'm finding a line that crosses through (2, -3). So, how can I do that? Does the original y equation factor into it somehow?


Sorry I ask so many questions here >.< I find the explanations very beneficial to my learning

 

[galactus]

You can use the point slope form of a line to find where the lines are tnagent to the parabola.

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

\(\displaystyle y'=m=2x+1\)

\(\displaystyle \underbrace{x^{2}+x}_{\text{y}}-\underbrace{(-3)}_{\text{y1}}=\overbrace{(2x+1)}^{\text{m}}(x-2)\)

Solve for x. There will be two solutions. Enter these into x^2+x to find their corresponding y values.

You can then use these, along with the given point, to find the equations of the two lines.

 

[mmm4444bot]

Sorry I ask so many questions here.

There is no need for you to be sorrowful.

As long as you follow the forum guidelines, there is no cap.

At least you post what you're thinking, with specific questions. The vast majority of people posting here don't say diddly. Most posters here simply dump their assignments, lay back, and wait for someone to "service " them.

My point: people like you are always welcome here; post as much as you like, without concern.

Cheers ~ Mark