So in a recent video, we learned all about tangent lines, and how to calculate the slope of a tangent line using an equation. Now recall that this is just a line that touches the curve of your function at exactly one point, that's what a tangent line is. And in this video, we're going to now learn how to solve some other problems associated with tangent lines, specifically finding their equations. Now if this sounds scary, don't sweat it, because it turns out the process for doing this is actually related to concepts that we've already learned up to this point. So, without further ado, let's just jump right into an example to see what these kinds of problems look like. So here we're asked to find the equation of the line tangent to this function, \( f(x) = 3x^2 - 4 \) at \( x = -2 \). So, how can we go about solving this? Well, what I'm going to do is do this by the steps. And our first step is going to be to plug our x-value of interest, which we call c, we discussed that in the previous video, into our function \( f(x) \) to get our y-value \( f(c) \). Now, I can see here that our c value is going to be -2 since that's the x-value of interest that we're looking at. So, since our x-value is -2, what we need to do is figure out what \( f(c) \) is, which is going to be \( f(-2) \). And \( f(-2) \), while plugging it into this function, will have \( 3 \times (-2)^2 - 4 \). Now, \( 3 \times (-2)^2 \) turns out to be 12, and 12 minus 4 is 8. So what we end up with is 8 for our \( f(c) \) value. So now that we found c and \( f(c) \), what we can do is move on to step 2, which is plugging things into this equation for the slope of a tangent line. Now we already figured out that the c here is -2, so this is going to stay consistent for this limit. But as for the rest of the equation, well, we need to plug our function in for \( f(x) \). \( f(x) \), we can see, is \( 3x^2 - 4 \). Now this whole thing is going to be minus \( f(c) \), which we just calculated to be 8. And all of this will be divided by \( x - c \), which is the same thing as \( x - (-2) \) or \( x + 2 \). So this is what we get for our equation. Now from here, what I'm going to do is simplify what we have on top since I can see I can take this -4 and subtract off 8. That's going to give me -12, so we're going to have \( 3x^2 - 12 \) all over \( x + 2 \). Now at this point, what I'm going to do is try applying my limit by pulling -2 in for x. But I notice that if I do that, I'm going to end up with -2 +2 on the bottom, which is 0. You cannot divide by 0. You cannot have 0 in the denominator of a fraction. So we're going to need to find some sort of other strategy here to evaluate this limit. Now what I'm going to do is take this 3 and factor it out of both these values because I can see that 3 would go on the outside, leaving us with just x squared there, and 12 divided by 3 is 4. So this whole thing will become \( x^2 - 4 \), and then all of that will be divided by \( x + 2 \). Now I still have this x plus 2 in the bottom, so I can't plug my -2 in yet, but what I notice is that we have \( x^2 - 4 \) on top. \( x^2 - 4 \) is a difference in squares, because we could also write this as \( x^2 - 2^2 \), because 2 squared is 4, all divided by \( x + 2 \). Since we have this difference in squares, this whole thing could be expanded out to be \( (x + 2)(x - 2) \), and again this is all divided by \( x + 2 \). Now from here, the \( x+2 \)s are going to cancel, meaning all that I'm going to have is \( 3 \times (x - 2) \). Now notice this x+2 in the denominator went away, so I can now plug my value of -2 in. So what we're going to have is \( 3 \times (-2 - 2) \). Now -2 minus 2 is -4, and 3 times -4 is -12. So that there is the slope of the tangent line. So notice that we've gone ahead and evaluated this limit, and we did have to factor and cancel. But that's very common when dealing with these types of limits. You're going to have to do that step. So now that we found what our slope of the tangent line is, we now need to figure out what the equation of our tangent line is. But we can actually do this because notice that we have an x and y value up here, and we have a slope. When we have these values, we can just use point slope form. So, that's the strategy for solving these types of problems. You first wanna calculate the slope, and then you wanna plug everything into point slope form to get your equation. So let's just go ahead and do this. So, we have \( y - \) our y-value, which is going to be \( y - 8 \). And then it's going to be equal to our slope, which we just calculated to be -12. This was our slope. And then it's going to be times \( x - \) our x-value, which is -2, or we could just write this as \( x + 2 \), because minus a negative would be plus. So now that we have everything plugged into point slope form, our last step is going to be to solve this equation for y. So let's go ahead and do that. Now the first thing I'm going to do is distribute this -12 into the parentheses. So what we're going to end up with is -12 times x, which is \( -12x \), and then -12 times 2, which is -24. Now from here, I'm going to add 8 on both sides of this equation, which will get the -8 and the positive 8 to cancel on the left. So that will give me \( y \) by itself which is what I'm looking for, and then we'll have \( -12x \), and then -24 +8 is -16. So \( y = -12x - 16 \) is the solution to this problem, and that's how you can find the equation of the tangent line. So, this is how you can solve these types of problems. Hope you found this video helpful, and let's try getting some more practice with this.