Find an equation for the line tangent to Y equal -5 -, 3 X squared at -3 negative 32. So the first thing we need to do is we need to find the slope at any given point of that equation Y equal -5 -, 3 X squared. So I'm going to do this in a very generic fashion versus at the specific point. So if I wanted to do it for any point on the curve, we'd use our limit as H goes to 0 of F of X + H -, F of X all over H. So the limit as H goes to zero -5 - 3. We're going to stick an X + H quantity squared there. Then we're going to subtract the F of X, which is -5 -, 3 X squared, and it's all going to go over H. Now we're going to do a bunch of algebra and simplification. So if we multiply the X + H ^2 quantity squared, we get X ^2 + 2 XH plus H ^2. If we distribute the -3 in the next line, we get negative three X ^2 -, 6 XH -3 H squared. Then the -5 and +5 will cancel the -3 X squared and the +3 X squared will cancel. So when our numerator gets left -6 XH -3 H squared all over H, if we factor an H out of that numerator, then we have an H over HH in the numerator, H in the denominator. That'll reduce to 1. So we'd get the limit as H goes to 0 of -6 X -3 H if H was zero -3 * 0 zero. So we get -6 X -6 X is the secant slope for absolutely any point on the given curve. Now the fact that they gave us the point -3 negative 32, we're going to actually find the sequence slope -6 X at that point. So I'm going to stick in -3 for the X and I get a slope of 18. Now we're given a point and we're given a slope. So we're going to use our point slope equation y - -32 equal 18 times the quantity X - -3. We do our algebra for a final answer of Y equal eighteen X + 22.