GMAT Math : Calculating the equation of a tangent line

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Example Questions

Example Question #1 : Tangent Lines

Determine the equation of the line tangent to the curve y=x^2 at the point (-1,1) ?

Possible Answers:

y=-2x-1

y=-2x+1

y=-x+2

y=x-2

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

Correct answer:

y=-2x-1

Explanation:

To find the equation of a line tangent to a curve at a certain point, we first need to find the slope of the curve at that point. To find the slope of a function at any point, we need its derivative:

$y=x^2\rightarrow y'=2x$

Now we can plug in the x value of the given point to find the slope of the function, and therefore the slope of tangent line, at that point:

y'=2x

$y'(-1)=2(-1)=-2\rightarrow m=-2$

Now that we have our slope, we can simply plug this value in with the given point to solve for the y intercept of the tangent line:

y=mx+b

1=(-2)(-1)+b

b=-1

We have calculated the slope of the tangent line and its y intercept, so the equation for the line tangent to the curve y=x^2 at the point (-1,1) in standard form is:

y=-2x-1

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Example Question #2 : Tangent Lines

Determine the equation of the line tangent to the following curve at the point (2,12) .

f(x)=3x^2+4x-8

Possible Answers:

y=-16x+20

y=10x-11

y=-10x+11

y=16x-20

y=16x+13

Correct answer:

y=16x-20

Explanation:

First we find the slope of the tangent line by taking the derivative of the function and plugging in the -value of the point where we want to know the slope:

f(x)=3x^2+4x-8

f'(x)=6x+4

f'(2)=6(2)+4=16

Now that we know the slope of the tangent line, we can plug it into the equation for a line along with the coordinates of the given point in order to calculate the -intercept:

y=mx+b

12=(16)(2)+b

b=-20

We now have and , so we can write the equation of the tangent line:

y=16x-20

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Example Question #3 : Tangent Lines

Find the equation of a line tangent to the curve $y=x^{2}-2x+1$ at the point (3,4) .

Possible Answers:

y=4x-8

None of the above equations

y=4x-4

y=4x+4

y=4x+8

Correct answer:

y=4x-8

Explanation:

To find the equation of a line tangent to the curve $y=x^{2}-2x+1$ at the point (3,4) , we must first find the slope of the curve at the point (3,4) by solving the derivative y'=2x-2 at that point:

$y'(3)=2(3)-2=6-2=4\rightarrow m=4$

Given the slope, we can now plug the given point and the slope at that point into the slope-intercept form of the tangent line y=mx+b and solve for the -intercept :

(4)=(4)(3)+b

4=12+b

-8=b

Given our slope, the chosen point, and the -intercept, we have the equation of our tangent line:

y=4x-8

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Example Question #4 : Tangent Lines

Find the equation of a line tangent to the curve $y=2x^{2}+5x+2$ at the point (2,-1) .

Possible Answers:

None of the above

y=27x-13

y=13x-27

y=27x+13

y=13x+27

Correct answer:

y=13x-27

Explanation:

To find the equation of a line tangent to the curve $y=2x^{2}+5x+2$ at the point (2,-1) , we must first find the slope of the curve at the point (2,-1) by solving the derivative y'=4x+5 at that point:

$y'(2)=4(2)+5=8+5=13\rightarrow m=13$

Given the slope, we can now plug the given point and the slope at that point into the slope-intercept form of the tangent line y=mx+b and solve for the -intercept :

(-1)=(13)(2)+b

-1=26+b

-27=b

Given our slope, the chosen point, and the -intercept, we have the equation of our tangent line:

y=13x-27

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Example Question #5 : Tangent Lines

Determine the equation of the tangent line to the following curve at the point (2,-7) :

f(x)=-2x^2+9x-17

Possible Answers:

y=-2x+3

y=x-7

y=x-9

y=x+9

y=-2x-3

Correct answer:

y=x-9

Explanation:

First find the slope of the tangent line by taking the derivative of the function and plugging in the x value of the given point to find the slope of the curve at that location:

f(x)=-2x^2+9x-17

f'(x)=-4x+9

f'(2)=-4(2)+9=1

So the slope of the tangent line to the curve at the given point is m=1 . The next step is to plug this slope into the formula for a line, along with the coordinates of the given point, to solve for the value of the y intercept of the tangent line:

y=mx+b

$-7=1(2)+b\rightarrow b=-9$

We now know the slope and y intercept of the tangent line, so we can write its equation as follows:

y=x-9

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Example Question #6 : Tangent Lines

Find the equation of a line tangent to the curve $y=4x^{2}-3x+7$ at the point (-2,-3) .

Possible Answers:

None of the above

y=19x+41

y=-19x-41

y=-19x+41

y=19x-41

Correct answer:

y=-19x-41

Explanation:

To find the equation of a line tangent to the curve $y=4x^{2}-3x+7$ at the point (-2,-3) , we must first find the slope of the curve at the point (-2,-3) by solving the derivative y'=8x-3 at that point:

$y'(-2)=8(-2)-3=-16-3=-19\rightarrow m=-19$

Given the slope, we can now plug the given point and the slope at that point into the slope-intercept form of the tangent line y=mx+b and solve for the -intercept :

(-3)=(-19)(-2)+b

-3=38+b

-41=b

Given our slope, the chosen point, and the -intercept, we have the equation of our tangent line:

y=-19x-41

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GMAT Math : Calculating the equation of a tangent line

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Tangent Lines

Example Question #2 : Tangent Lines

Example Question #3 : Tangent Lines

Example Question #4 : Tangent Lines

Example Question #5 : Tangent Lines

Example Question #6 : Tangent Lines

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