Recall that the slope of a tangent line of a point is the same as the derivative of the function at that point. So, to find an equation of the tangent line at a certain point, calculate the derivative of the function at that point to find its slope. Then, using the point at which you wound the tangent line, you can use point-slope form to find the y-intercept. Remember that point-slope form is y-y1=m(x-x1), where (x1,y1) is the point where you are finding the tangent line, and m is the slope calculated by using the derivative.
So one problem that you might see on your homework is; find the equation of a line tangent to a curve at a certain point. This is something you can do with the derivative, because the slope of a tangent line at a given point comes from the derivative. But just remember that when you want to find the equation of a line, you need two things. You need a point that the line passes through, and the slope.
So let's take a look at this example of a problem here. It says write an equation of the line tangent to the graph of. Then I have this big polynomial function f(x) equals 2x³ minus 5x² plus 3x minus 5. The point of tangency is going to be at x equals 1. So first I want to find both the x and the y coordinate of the point of tangency. So this is the x coordinate. The y coordinate is going to be f(1). So that's 2 times 1³, 2 times 1minus 5 times 1², 5 times 1 plus 3 times 1 minus 5. So I have 2 minus 5, -3, plus 3, 0, -5. That means that the point of tangency is going to be 1,-5.
So all I have to do now is find the slope. The slope comes from the derivative.
The derivative of this thing is going to be, the derivative with respect to x of all of this 2x³ minus 5x² plus 3x minus 5. So first, I want to break this part using the sum rule. So I have the derivative with respect to x of 2x³ plus the derivative with respect to x, of -5x², plus the derivative with respect to x of 3x minus 5. Three pieces.
Then I'm going to use the constant multiple rule to pull the constants out. So you get 2 times the derivative with respect to x of x³, plus -5, this constant, times the derivative of x². I don't really need to pull anything out here. This is just a linear function. I know the derivative of linear function is going to be the slope of that function 3.
Just going through, let me continue this up here. So I've got 2 times the derivative of x³. That's going to be 3x². 2 times 3x² plus -5 times the derivative of x², 2x. So -5 times 2x plus 3. So this is f'(x).
Let me just simplify that. 6x² minus 10x plus 3. So what you need is the slope of the tangent line. You need the slope specifically at x equals 1. So I'm going to need to calculate f'(1). That's going to be 6 times 1 minus 10 times 1 plus 3. So 6 minus 10, -4 plus 3, -1. That's your slope.
Now finally use the point slope formula to find the equation of the line. So you have y minus, don't forget the coordinates of our point of tangency 1, -5. So y minus -5, y plus 5 equals the slope -1 times x minus 1. This is actually a perfectly good answer in point slope form. But if your teacher wants slope-intercept form, you'd have -x plus 1, and then subtract 5 from that minus 4. Y equals -x minus 4. That's the equation of the line tangent to our polynomial at x equals 1.