### All Calculus 1 Resources

## Example Questions

### Example Question #1 : Lines

What is the equation of the line tangent to f(x) = 4x^{3} – 2x^{2} + 4 at x = 5?

**Possible Answers:**

y = 85x + 24

None of the other answers

y = 220x – 550

y = 44x + 245

y = 280x – 946

**Correct answer:**

y = 280x – 946

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

f'(x) = 12x^{2} – 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 * 5^{3} – 2 * 5^{2} + 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 #1 : Lines

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

**Possible Answers:**

**Correct answer:**

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 #231 : Graphing Functions

Find the equation of the line tangent to at .

**Possible Answers:**

**Correct answer:**

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 #232 : Graphing Functions

Find the equation of the line tangent to at .

**Possible Answers:**

**Correct answer:**

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 #233 : Graphing Functions

Find the equation of the line tangent to at .

**Possible Answers:**

**Correct answer:**

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 #234 : Graphing Functions

A function, , is given by

.

Find the line tangent to at .

**Possible Answers:**

**Correct answer:**

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 #235 : Graphing Functions

Let .

Find the equation for a line tangent to when .

**Possible Answers:**

**Correct answer:**

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 #236 : Graphing Functions

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

**Possible Answers:**

**Correct answer:**

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 #237 : Graphing Functions

Find the equation of the tangent line, where

, at .

**Possible Answers:**

**Correct answer:**

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 #238 : Graphing Functions

Find the equation of the tangent line of

, at .

**Possible Answers:**

**Correct answer:**

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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