Calculus 1 : How to find equation of line by graphing functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

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Example Question #1 : Equation Of Line

What is the equation of the line tangent to f(x) = 4x3 – 2x2 + 4 at x = 5?

Possible Answers:

y = 220x – 550

None of the other answers

y = 280x – 946

y = 85x + 24

y = 44x + 245

Correct answer:

y = 280x – 946

Explanation:

First, take the derivative of f(x). This is very easy:

f'(x) = 12x2 – 4x

Now, the slope of the tangent line at x = 5 is f'(5). Evaluated, this is: f'(5) = 12 * 5 * 5 – 4 * 5 = 300 – 20 = 280

Now, we must find the intersection point on the original line:

f(5) = 4 * 53 – 2 * 52 + 4 = 500 – 50 + 4 = 454

Therefore, the point of tangent intersection is (5,454).

Using the point-slope form of linear equations, we can find the line:

(y – 454) = 280(x – 5)

y – 454 = 280x – 1400

y = 280x – 946

Example Question #4 : Lines

Find the equation of the line tangent to  at the point .

Possible Answers:

Correct answer:

Explanation:

The equation of the tangent line will have the form , where  is the slope of the line and .  

To find the slope, we need to evaluate the derivative at :

 

Now that we have the slope, we can determine the equation of the tangent line using the point-slope formula:

 

Example Question #5 : Lines

Find the equation of the line tangent to  at .

Possible Answers:

Correct answer:

Explanation:

To find the equation of the line at that point, you need two things: the slope at that point and the y adjustment of the function, in the form , where m is the slope and  is the y adjustment. To get the slope, find the derivative of  and plug in the desired point  for , giving us an answer of  for the slope.

To find the y adjustment pick a point 0 in the original function. For simplicity, let's plug in , which gives us a y of 1, so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is

Now you simply solve for , which is .

Final equation of the line tangent to  at  is 

Example Question #1 : Equation Of Line

Find the equation of the line tangent to  at .

Possible Answers:

Correct answer:

Explanation:

To get the slope, find the derivative of  and plug in the desired point  for , giving us an answer of  for the slope.

Remember that the derivative of .

 

To find the  adjustment pick a point  (for example) in the original  function. For simplicity, let's plug in , which gives us a  of , so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is

The coefficient in front of the  is the slope.

Now you simply solve for , which is .

Final equation of the line tangent to  at is .

Example Question #2 : Equation Of Line

Find the equation of the line tangent to  at  .

Possible Answers:

Correct answer:

Explanation:

The equation of the tangent line is  To find , the slope, calculate the derivative and plug in the desired point.

The next step is to choose a coordinate on the original  function. We can choose any  value and calculate its  value.

Let's choose .

The  value at this point is .

Plugging in those values we can solve for .

Solving for  we get =.

Example Question #8 : Lines

A function, , is given by

.

Find the line tangent to  at .

Possible Answers:

Correct answer:

Explanation:

First we need to find the slope of  at . To do this we need the derivative of . To take the derivative we need to use the power rule for the first term and recognize that the derivative of sine is cosine.

At,

 

Now we need to know

.

Now we have a slope,  and a point 

so we can use the point-slope formula to find the equation of the line.

Plugging in and rearranging we find

.

Example Question #9 : Lines

Let 

Find the equation for a line tangent to  when .

Possible Answers:

Correct answer:

Explanation:

First, evaluate  when .

Thus, we need a line that contains the point 

Next, find the derivative of  and evaluate it at .

To find the derivative we will use the power rule,

.

This indicates that we need a line with a slope of 8.

In point-slope form, , a line with the point  and a slope of 8 will be:

Example Question #3 : Equation Of Line

What is the equation of the line tangent to  at ? Round to the nearest hundreth. 

Possible Answers:

Correct answer:

Explanation:

The tangent line to  at  must have the same slope as .

Applying the chain rule we get

.

Therefore the slope of the line is, 

.

In addition, the tangent line touches the graph of  at . Since , the point  lies on the line.

Plugging in the slope and point we get .

Example Question #1 : Equation Of Line

Find the equation of the tangent line, where

, at .

Possible Answers:

Correct answer:

Explanation:

In order to find the equation of the tangent line at , we first find the slope.

To do this we need to find .

Since we have found , now we simply plug in 1.

Now we need to plug in 1, into , to find a point that the tangent line touches.

Now we can use point-slope form to figure out what the equation of the tangent line is at .

Remember that point-slope for is

where  and  is the point where the tangent line touches , and  is the slope of the tangent line.

In our case, , and .

Thus our tangent line equation at  is

.

 

Example Question #2 : Equation Of Line

Find the equation of the tangent line of 

, at .

Possible Answers:

Correct answer:

Explanation:

In order to find the equation of the tangent line at , we first find the slope.

To do this we need to find  using the power rule .

Since we have found , now we simply plug in 1.

Now we need to plug in 1, into , to find a point that the tangent line touches.

Now we can use point-slope form to figure out what the equation of the tangent line is at .

Remember that point-slope for is

where  and  is the point where the tangent line touches , and  is the slope of the tangent line.

In our case, , and .

Thus our tangent line equation at  is

.

 

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