Find the slope of the curve Y equal X ^2 -, 2 X -3 at the point P2, -3 by finding the limit of the secant slopes through point P Find an equation the tangent line to the curve at P2 negative 3. So I'm going to use the official definition to find the slope of the curve. The limit is H goes to 0 of F of X + H -, F of X all over H. So the limit is H goes to 0. Now I'm going to do this generically for any point. The reason to do it generically for any point is then once we come up with an equation, we can easily substitute to find the information for any point, not just for one single given point. So I'm going to stick an X + H into the original equation. So we're going to get X + H ^2 -, 2 times the quantity X + H - 3 minus the original F of X equation, which was X ^2 -, 2 X -3 all over H. Now with some algebra, we're going to distribute and combine like terms foil X ^2 + 2 XH plus H ^2 -, 2 X -2 H -3 -, X ^2 + 2 X +3 all over H. We can see that quite a few things cancel, leaving us two XH plus H ^2 -, 2 H over H to factor out an H so that the H in the numerator and the H in the denominator cancel, which will give me two X + H - 2 as the limit. The limit as H goes to 0. So we're going to stick zero literally in for the H now because it's not going to make it undefined. And we're going to come up with an equation of two X -, 2. So now if we wanted to do it at the .2 negative 3:00, we'd literally stick 2IN for our X. So 2 * 2 - 2 is going to give us 2, so the slope of the curve at that specific point is 2. Then we're going to just use point slope form y - -3 equal two times the quantity X - 2 Y equal two X - 7.