 In this video, we provide the solution to question number 8 for practice exam number 3 for math 1220, in which case we have to find the equation of the tangent line given by the parametric curve x equals t cube plus 6t plus 1 and y equals 2t minus t squared at the point where t equals negative 1. We need to make sure we write it in slope intercept form. So in the end we want something to look like y equals mx plus b, that's what our goal is, but the standard equation for a tangent line is going to be y minus, we have our point of tangency for which we're going to say y of t at that moment. This equals then the derivative dy over dx evaluated at some value for t and times that by x minus x evaluated at t. So we want to look for those things. So our t value here is negative 1, so we need to compute what is x at negative 1. This is going to equal negative 1 cubed plus 6 times negative 1 plus 1. Let's try to simplify that. Negative 1 cubed is negative 1. 6 times negative 1 is negative 6 plus 1. The plus 1 and minus 1 cancel out and so we're left with x is negative 6 in this situation. We need to compute y at negative 1. We plug that into the formula. We're going to get 2 times negative 1 minus negative 1 squared. 2 times negative 1 is negative 2. Negative 1 squared is positive 1, but you subtract from it 1, so you end up with negative 3. So what do we observe so far? We're going to get a y minus the y-coordinate, so we get 3 plus, y plus 3 there. The derivative, we don't know it yet. We're going to come back to that one. Leave a space open for that. Then we're going to get x minus the x-coordinate, so we get x minus the negative 6 as x plus 6. So we have the start of our function, tangent line here. What needs to still be computed is the derivative dy over dx, excuse me, for which by definition, the derivative of y with respect to x, not by definition, I should say by the chain rule here, the derivative of y with respect to x is the derivative of y with respect to t divided by the derivative of x with respect to t, like so. So we take the derivative of y with respect to t. We're going to get 2 minus 2t over the derivative of x with respect to t, which is 3t squared plus 6 for which we could try to simplify this, but admittedly, like we said, we only need to evaluate this when t equals negative 1. That's really what we care about when t equals negative 1. So instead of simplifying the derivative, I'm just going to plug in negative 1 at this moment. So we get 2 minus 2 times negative 1. On top, on the bottom, we get 3 times negative 1 squared plus 6. So on the top, you get a negative 2 times a negative 1, which is a positive 2. On the bottom, negative 1 squared is a positive 1, so you get 3 plus 6 there. So on the top, you get 4. On the bottom, you get 9. And so that is then the slope of this thing we're going to put in there. So we end up with a 4 ninths. We do need to put this into slope intercept form, so I'm going to distribute that 4 ninths there. So we get y plus 3 is equal to 4 ninths x plus 4 ninths times 6. Do notice that 9 and 6 are both divisible by 3. So you get 3 there, 2 there. So you can actually rewrite that fraction. So we're going to end up with y equals 4 ninths x plus, when you rewrite that fraction, you're going to get 4 times 2, which is 8 over 3. I'm going to move this plus 3 to the other side by subtraction. So I'm going to rewrite that as negative 9 over 3. And then combining like terms there, you end up with your slope intercept form version of the tangent line, y equals 4 ninths x minus 1 third. And that is our final answer there.