 In this video we will discuss the solution for question 9 on the practice final exam for math 1220. In this question we are asked to find the equation of the tangent to the curve x equals t cubed plus 6t plus 1 and y equals 2t minus t squared when t equals negative 1. And we are going to report this tangent line in slope intercept form. We have to find the equation of the tangent line and so as a reminder that equation will have the form. You get y minus, well the function, the point of tangency y1 here. This is going to equal dy dx times x minus x1. So x1 and y1 are point of tangency. And so that comes into play when t equals negative 1. So that's something we can find out pretty quickly here. So when you take t equals negative 1x1 will equal negative 1 cubed plus 6 times negative 1 plus 1. Simplifying that creature we get negative 1 minus 6 plus 1. Though ones cancel out and we are left with a negative 6 for the x-coordinate. And then for the y-coordinate y1 we plug in negative 1. We are going to get 2 times negative 1 minus negative 1 squared. So we get negative 2 minus 1 which is a negative 3. And so now we have our point negative 6 comma negative 3. This is the point of tangency that we are going to need. And so kind of preparing for the final answer here we are going to get y plus 3 because it's a minus negative 3 there. So this equals something times x plus 6. And so of course we do have to fill in that something in just a moment. And so that then brings us to the derivative calculation. dy dx are which we often abbreviate as y prime. For a parametric curve this is calculated as dy over dt divided by dx over dt. So we calculate the derivative of y and x with respect to t and take their ratios. The derivative of y with respect to t just by the usual power rule will be 2 minus 2t. The derivative of x by the power rule with respect to t would be 3t squared plus 6. And then we evaluate this when t equals negative 1. That's the value we need to do in which case that gives us 2 minus 2 times negative 1 on top. We have 3 times negative 1 squared plus 6 on the bottom. In the numerator you're going to end up with 2 plus 2. In the denominator you're going to end up with 3 plus 6. So we end up with 4 over 9 four ninths. That sounds pretty good for me. And so we plug that into our point slope form. Now the instructions did specify to write the final answer in slope intercept form. So that just means we have to solve for y. Distribute the four ninths through. We'll get y plus 3 equals four ninths x. Plus you're going to get 4 times 6 over 9. I always think it's better to reduce before you multiply, expand it out, right? Because 6 and 9 have a common factor of 3. So 3 goes into 6 two times. It goes into 9 three times. And so we're going to have an 8 thirds there. So y equals 4 over 9x plus 8 thirds minus 3. So we subtracted the 3 from both sides. Now in order to find a common denominator we should write this as 9 over 3. So negative 9 over 3. And therefore combining the fractions there you get y equals 4 ninths x minus 1 third. And so this right here gives us the slope. It gives us the equation of the tangent line written in slope intercept form.