Brightstorm is like having a personal tutor for every subject

See what all the buzz is about

Check it out
Don't Stop Learning Now!

Gain access to 3,500 HD videos.

Convince me!

Watch 1 minute preview of this video

or

Get Immediate Access with 1 week FREE trial
Your video will begin after this quick intro to Brightstorm.

Derivatives of Polynomial Functions - Problem 2 3,110 views

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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.