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

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Example Questions

Example Question #1 : Lines

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

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

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Example Question #1 : Lines

Find the equation of the line tangent to y=x^2-x+3 at the point (1,3) .

Possible Answers:

y=0.5x+2

y=0.5x+1

y=2x+2

y=x+2

Correct answer:

y=x+2

Explanation:

The equation of the tangent line will have the form $y-y_{0}=m(x-x_{0})$ , where is the slope of the line and $(x_{0},y_{0})=(1,3)$ .

To find the slope, we need to evaluate the derivative at x=1 :

$\frac{dy}{dx}=\frac{d(x^2-x+3)}{dx}=2x-1\Rightarrow m=f'(1)=2*1-1=1$

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

$y-y_{0}=m(x-x_{0})\Rightarrow y-3=1(x-1)\Rightarrow y=x+2.$

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Example Question #1 : How To Find Equation Of Line By Graphing Functions

Find the equation of the line tangent to f(x) at x=0 .

$f(x)=e^{3x} +x^2$

Possible Answers:

y=3x+2

y=3x+1

y=x^2 +2

y=3

$y=3e^{3x} +2x$

Correct answer:

y=3x+1

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 y=mx+b , where m is the slope and is the y adjustment. To get the slope, find the derivative of f(x) and plug in the desired point for , giving us an answer of for the slope.

$f'(x)=3e^{3x} + 2x$

f'(0)=3

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

1=(3)(0)+b

Now you simply solve for , which is .

Final equation of the line tangent to f(x) at x=0 is y=3x+1

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Example Question #1 : Lines

Find the equation of the line tangent to f(x) at x=2 .

f(x)= ln(2x+1) +3

Possible Answers:

y=3x

y=2x +3

$y=\frac{2}{5}x$

$y=\frac{2}{5}x +3$

$y=\frac{2}{3}x$

Correct answer:

$y=\frac{2}{5}x +3$

Explanation:

To get the slope, find the derivative of f(x) and plug in the desired point for , giving us an answer of $\frac{2}{5}$ for the slope.

$f'(x)=\frac{2}{2x+1}$

$f'(2)=\frac{2}{5}$

Remember that the derivative of $ln(u) = \frac{u'}{u}$ .

To find the adjustment pick a point (for example) in the original f(x) function. For simplicity, let's plug in x=0 , which gives us a of , so an easy point is (0,3) . Next plug in those values into the equation of a line, y=mx + b . The new equation with all parameters plugged in is

$3=\frac{2}{5}(0)+b$

The coefficient in front of the is the slope.

Now you simply solve for , which is .

Final equation of the line tangent to f(x) at x=2 is $y=\frac{2}{5}x +3$ .

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Example Question #1 : Equation Of Line

Find the equation of the line tangent to y=sin(2x) at $\frac{\pi }{2}$ .

Possible Answers:

y=2cos(2x)

y=-2x+1

$y=-2x+\pi$

y=0

$y=2x-\pi$

Correct answer:

$y=-2x+\pi$

Explanation:

The equation of the tangent line is y=mx+b. To find , the slope, calculate the derivative and plug in the desired point.

y'=2cos(2x)

$y'=2cos\left(2\cdot \frac{\pi}{2}\right)$

$y'=2cos(\pi)$

$y'\left(\frac{\pi}{2} \right )=-2$

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

Let's choose $\frac{\pi }{2}$ .

$y=sin\left(2\cdot \frac{\pi}{2}\right)=sin(\pi)=0$

The value at this point is .

Plugging in those values we can solve for .

$0=(-2)\left(\frac{\pi }{2}\right)+b$

Solving for we get = $\pi$ .

$y=-2x+\pi$

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Example Question #2 : Equation Of Line

A function, , is given by

f(x)=x^2-sin(x) .

Find the line tangent to at $x=\frac{\pi}{2}$ .

Possible Answers:

$y=\frac{\pi}{2} x-\frac{\pi^2}{4}$

$y=\pi x-1$

$y=\pi x-\frac{\pi^2}{4}-1$

$y=2\pi x-\frac{\pi^2}{2}+1$

$y=\pi x-\frac{\pi^2}{2}$

Correct answer:

$y=\pi x-\frac{\pi^2}{4}-1$

Explanation:

First we need to find the slope of at $x=\frac{\pi}{2}$ . 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.

$\frac{df}{dx}=2x-cos(x)$

At,

$x=\frac{\pi}{2}, \frac{df}{dx}=\pi$

Now we need to know

$f\left(\frac{\pi}{2}\right)=\frac{\pi^2}{4}-1$ .

Now we have a slope, $m=\pi$ and a point $(x_1,y_1)=\left(\frac{\pi}{2},\frac{\pi^2}{4}-1\right)$

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

y-y_1=m(x-x_1)

Plugging in and rearranging we find

$y=\pi x-\frac{\pi^2}{4}-1$ .

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Example Question #4 : How To Find Equation Of Line By Graphing Functions

Let f(x)=2x^4+e^x-ex .

Find the equation for a line tangent to f(x) when x=1 .

Possible Answers:

y-1=2(x-2)

y+8=2(x+1)

y-8=2(x-1)

y-2=8(x-1)

8y=2(x-1)

Correct answer:

y-2=8(x-1)

Explanation:

First, evaluate f(x) when x=1 .

$f(1)=2(1)^4+e^1-e(1)=2+0\textbf{=2}$

Thus, we need a line that contains the point (1,2)

Next, find the derivative of f(x) and evaluate it at x=1 .

To find the derivative we will use the power rule,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

f'(x)=8x^3+e^x-e

f'(1)=8(1)^3+e^1-e=8+0=8

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

In point-slope form, $y-y_{1}=m(x-x_{1})$ , a line with the point (1,2) and a slope of 8 will be:

y-2=8(x-1)

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Example Question #2 : Lines

What is the equation of the line tangent to $f(x)=\arctan (1-x^3)$ at x=-1 ? Round to the nearest hundreth.

Possible Answers:

y=-0.6

y=-0.6x

y=0.2x+1.31

y=0.2x

y=-0.6x+0.51

Correct answer:

y=-0.6x+0.51

Explanation:

The tangent line to f(x) at x=-1 must have the same slope as f(x) .

Applying the chain rule we get

$f'(x)=\frac{1}{1+(1-x^3)^2}(-3x^2)$ .

Therefore the slope of the line is,

$f'(-1)=-\frac{3}{5}=-0.6$ .

In addition, the tangent line touches the graph of f(x) at x=-1 . Since f(-1)=1.11 , the point (-1, 1.11) lies on the line.

Plugging in the slope and point we get y=-0.6x+0.51 .

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Example Question #231 : Graphing Functions

Find the equation of the tangent line, where

f(x)=x^4-x^2+x-5 , at x=1 .

Possible Answers:

y=3x+1

y=3x-7

y=3x+7

y=3x-1

y=x

Correct answer:

y=3x-7

Explanation:

In order to find the equation of the tangent line at x=1 , we first find the slope.

To do this we need to find f'(x) .

f(x)=x^4-x^2+x-5

f'(x)=4x^3-2x+1

Since we have found f'(x) , now we simply plug in 1.

f'(1)=4(1)^3-2(1)+1=3

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

f(1)=(1)^4-(1)^2+1-5=-4

Now we can use point-slope form to figure out what the equation of the tangent line is at x=1 .

Remember that point-slope for is

y-y_0=m(x-x_0)

where x_0 and y_0 is the point where the tangent line touches f(x) , and is the slope of the tangent line.

In our case, x_0=1 , y_0=-4 , and m=3 .

y+4=3(x-1)

y+4=3x-3

Thus our tangent line equation at x=1 is

y=3x-7 .

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Example Question #3 : Equation Of Line

Find the equation of the tangent line of

f(x)=x^3+x , at x=1 .

Possible Answers:

y=2x+7

y=4x-2

y=4x+2

y=4x

y=-4x-2

Correct answer:

y=4x-2

Explanation:

In order to find the equation of the tangent line at x=1 , we first find the slope.

To do this we need to find f'(x) using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

f(x)=x^3+x

f'(x)=3x^2+1

Since we have found f'(x) , now we simply plug in 1.

f'(1)=3(1)^2+1=4

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

f(1)=(1)^3+1=2

Now we can use point-slope form to figure out what the equation of the tangent line is at x=1 .

Remember that point-slope for is

y-y_0=m(x-x_0)

where x_0 and y_0 is the point where the tangent line touches f(x) , and is the slope of the tangent line.

In our case, x_0=1 , y_0=2 , and m=4 .

y-2=4(x-1)

y-2=4x-4

Thus our tangent line equation at x=1 is

y=4x-2 .

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

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : Lines

Example Question #1 : Lines

Example Question #1 : How To Find Equation Of Line By Graphing Functions

Example Question #1 : Lines

Example Question #1 : Equation Of Line

Example Question #2 : Equation Of Line

Example Question #4 : How To Find Equation Of Line By Graphing Functions

Example Question #2 : Lines

Example Question #231 : Graphing Functions

Example Question #3 : Equation Of Line

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