GMAT Math : Coordinate Geometry

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Example Questions

Example Question #2 : Tangent Lines

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

Possible Answers:

None of the above equations

y=4x+8

y=4x-4

y=4x-8

y=4x+4

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 #3 : Calculating The Equation Of A Tangent Line

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

Possible Answers:

y=27x+13

y=13x+27

y=27x-13

y=13x-27

None of the above

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 #1 : 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-9

y=x+9

y=x-7

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 #2 : Calculating The Equation Of A Tangent Line

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

Possible Answers:

y=19x+41

y=-19x+41

y=-19x-41

None of the above

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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Example Question #1 : Calculating The Slope Of A Tangent Line

Find the slope of the line tangent to f(x) at x=3 .

$\small f(x)=x^2-6x+9$

Possible Answers:

Correct answer:

Explanation:

f(x) in this case gives us a parabola. If we factor the f(x) we get:

$\small f(x)=(x-3)(x-3)$

This equation has one zero and that is at x=3. This also means we have a minimum at x=3.

The slope of a line tangent to a minimum or maximum is always 0, so our slope is zero.

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Example Question #1 : Calculating The Slope Of A Tangent Line

Calculate the slope of the tangent line to the following curve at the point (1,4) .

f(x)=-2x^2+13x-7

Possible Answers:

Correct answer:

Explanation:

The slope of the tangent line to a curve at any point is simply the slope of the curve at that point. To find the slope of the function at any point, we take the derivative:

f(x)=-2x^2+13x-7

f'(x)=-4x+13

Now we can plug in the x value of the given point, (1,4) , which gives us the slope of the tangent line to the curve at that point:

f'(x)=-4x+13

f'(1)=-4(1)+13=9

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Example Question #3 : Tangent Lines

Calculate the slope of the line tangent to the curve y=3x^2-5x+7 at the point (2,15) .

Possible Answers:

Correct answer:

Explanation:

To calculate the slope of a tangent line at a particular point, we need to know the slope of the curve at that point. In order to find the slope of a curve at any point, we need to calculate its derivative:

y=3x^2-5x+7

y'=6x-5

The derivative describes the slope of the curve at any point, so we need to plug in the value of the given point (2,15) to find the slope of the tangent line at that location:

y'=6x-5

y'(2)=6(2)-5=7

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Example Question #1 : Calculating The Slope Of A Tangent Line

Calculate the slope of the line tangent to the following curve at x=1 :

f(x)=6x^5-7x^4+2x^3

Possible Answers:

Correct answer:

Explanation:

First find the derivative of the function, and then plug in the given x coordinate, which will give the slope of the function at that location. The slope of the tangent line to a curve at a given location is equal to the slope of the function at that location:

f(x)=6x^5-7x^4+2x^3

f'(x)=30x^4-28x^3+6x^2

f'(1)=30(1)^4-28(1)^3+6(1)^2=8

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Example Question #1 : Calculating The Slope Of A Tangent Line

Calculate the slope of the line tangent to the following curve at x=-2 :

f(x)=4x^2+12x-7

Possible Answers:

Correct answer:

Explanation:

f(x)=4x^2+12x-7

f'(x)=8x+12

f'(-2)=8(-2)+12=-4

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Example Question #1 : Calculating The Slope Of A Tangent Line

Calculate the slope of the line tangent to the following curve at x=0 :

$f(x)=e^{5x}-ln(15x+5)$

Possible Answers:

Correct answer:

Explanation:

First find the derivative of the function. The slope of the tangent line to a curve at a given point is equal to the slope of the function at that point, and the derivative of a function tells us its slope at any point. Plugging the given point into the equation for the derivative, we can calculate the slope of the function, and therefore the slope of the tangent line, at that point:

$f(x)=e^{5x}-ln(15x+5)$

$f'(x)=5e^{5x}-\frac{15}{15x+5}$

$f'(0)=5e^{0}-\frac{15}{15(0)+5}=5-3=2$

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GMAT Math : Coordinate Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #2 : Tangent Lines

Example Question #3 : Calculating The Equation Of A Tangent Line

Example Question #1 : Tangent Lines

Example Question #2 : Calculating The Equation Of A Tangent Line

Example Question #1 : Calculating The Slope Of A Tangent Line

Example Question #1 : Calculating The Slope Of A Tangent Line

Example Question #3 : Tangent Lines

Example Question #1 : Calculating The Slope Of A Tangent Line

Example Question #1 : Calculating The Slope Of A Tangent Line

Example Question #1 : Calculating The Slope Of A Tangent Line

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