# Calculus 1 : How to find slope by graphing functions

## Example Questions

← Previous 1 3

### Example Question #1 : How To Find Slope By Graphing Functions

What is the slope of the tangent line of f(x) = 3x4 – 5x3 – 4x at x = 40?

743,996

768,000

684,910

331,841

743,996

Explanation:

The first derivative is easy:

f'(x) = 12x3 – 15x2 – 4

The slope of the tangent line is found by calculating f'(40) = 12 * 403 – 15 * 402 – 4 = 768,000 – 24,000 – 4 = 743,996

### Example Question #2 : Slope

Find the slope of the line tangent to  when  is equal to .

Explanation:

To find the slope of a tangent line, we need to find the first derivative of the function at that point. In other words, we need y'(6).

Taking the first derivative using the Power Rule  we get the following.

Substituting in 6 for b and solving we get:

.

### Example Question #1 : How To Find Slope By Graphing Functions

Find function which gives the slope of the line tangent to .

Explanation:

To find the slope of a tangent line, we need the first derivative.

Recall that to find the first derivative of a polynomial, we need to decrease each exponent by one and multiply by the original number.

### Example Question #1 : Slope

Find the slope of the line tangent to  at .

Explanation:

The slope of the tangent line can be found easily via derivatives. To find the slope of the tangent line at s=16, find b'(16) using the power rule on each term which states:

Applying this rule we get:

Therefore, the slope we are looking for is 454.

### Example Question #2 : Slope

Find the slope of  at .

Explanation:

To find the slope of the line at that point, find the derivative of f(x) and plug in that point.

Remember that the derivative of  and the derivative of

Now plug in

### Example Question #6 : Slope

Find the slope of at  given . Assume the integration constant is zero.

Explanation:

The first step here is to integrate  in order to get .

Here the problem tells us that the integration constant , so

Plug in  here

### Example Question #7 : Slope

Consider the curve

.

What is the slope of this curve at ?

Explanation:

The slope of a curve at any point is equal to the derivative of the curve at that point.

Remembering that the derivative of  and using the power rule on the second term we find the derivative to be:

.

Pluggin in  we find that the slope is .

### Example Question #1 : Slope

Find the line tangent to  at .

Explanation:

Find the line tangent to  at .

First, we find :

Next, we find the derivative:

Therefore, the slope at  is:

.

Using point-slope form, we can write the tangent line:

Simplifying this gives us:

### Example Question #9 : Slope

An isosceles triangle has one point at , one point at  and one point on the -axis. What is the slope of the line between the point on the -axis and ?

Explanation:

The other point of the triangle must be at  as it must be equidistant from the other two points of the triangle. Since all points on the y-axis are  units away from the other points in the  direction, the third point must be equidistant in the  direction from both  and . The distance between these points is , so the third point must have a y-value of . The third point is now at  so the slope of the line from  to  is as follows.

### Example Question #10 : Slope

What is the slope of the line tangent to the graph of  at ?

Explanation:

We must take the derivative of the function using the chain rule yielding .

The chain rule is .

Also remember that the derivative of  is .

Applying these rules we get the following.

Plugging in the value for  we get  which is .

← Previous 1 3