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GRE Subject Test: Math : GRE Subject Test: Math

Study concepts, example questions & explanations for GRE Subject Test: Math

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Example Questions

Example Question #1 : Finding Equations Of Lines

Find the equation of a line in slope-intercept form that passes though points (-1,3) and (-2,5) .

Possible Answers:

y = -2x - 1

y = 2x -1

y = -2x + 1

y = 2x + 1

Correct answer:

y = -2x + 1

Explanation:

To find the equation of a line that passes through (-1,3) and (-2,5) , first find the slope using this formula:

$m = \frac{y_{2} -y_{1}}{x_{2}-x_{1}}$

$m = \frac{5-3}{-2-(-1)} = \frac{2}{-1}$

m = -2

Using one set of points or coordinates and the value of slope, plug these values into:

y=mx+b

Either set of points will give you the same equation of the line.

Coordinates (-1,3) .

y-3 = -2(x - (-1)

y-3 = -2(x+1)

Distribute the to what is inside the parenthesis.

$-2\times x = -2x$

$-2 \times 1 = -2$

y-3 = -2x-2

Add to both sides of the equation.

y-3+3 = -2x-2+3

The equation of the line in slope-intercept form y=mx+b is:

y = -2x +1

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Example Question #1 : Finding Equations Of Lines

A line has a slope of $m=-\frac{1}{4},$ and goes through point (0,8). What is the equation of the line in slope-intercept form?

Possible Answers:

$y = -\frac{1}{4} -8$

$y = \frac{1}{4} +8$

$y = \frac{1}{4} - 8$

$y = -\frac{1}{4} +8$

Correct answer:

$y = -\frac{1}{4} +8$

Explanation:

The slope-intercept form is y = mx+b , where represents the slope, and represent the points, and is the y-intercept or the value of when x=0.

The slope or $\frac{rise}{run}$ has been given as $m=-\frac{1}{4}$ .

The points that this equation of the line passes through are (0,8) and

b=8 because based on the points, when x=0,y=8. That is the y-intercept.

The equation of the line in slope-intercept form y=mx+b is:

$y=-\frac{1}{4} x+8$

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Example Question #2 : Finding Equations Of Lines

Find the equation of a line in slope-intercept form that passes through points (-5,-3) and has a slope of m=-2.

Possible Answers:

y = -2x-13

y = 2x+13

y = 2x-13

y = -2x+13

Correct answer:

y = -2x-13

Explanation:

To find the equation of a line given points (-5, -3) and m = -2, use the point- slope formula:

$y-y_{1} = m(x -x_{1})$

y -(-3) = -2 (x--5)

y +3 = -2(x+5)

Distribute the to what is inside the parenthesis.

-2(x) = -2x

$-2\times 5 = -10$

y + 3 = -2x-10

Subtract from both sides of the equation.

y+3-3 = -2x-10-3

-10-3 = -13 When the sign is the same for both integers, add.

The equation of the line in slope-intercept form y=mx+b is:

y=-2x-13

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Example Question #1 : Finding Equations Of Lines

Find the equation of the line with points (1,6) (3, -2)

Possible Answers:

y=1/4-10

y=-4x+10

y=4-25/4

y=-1/4x+25/4

y=10x-4

Correct answer:

y=-4x+10

Explanation:

y=mx+b

To solve for you must use the equation $(y^{1}-y^{2})/(x^{1}-x^{2})$

(6-(-2))/(1-3) = -8/2 = -4

y=-4x+b

To solve for be we must plug in one of the points

6=-4(1)+b

simplify

6=-4+b

Add to both sides

10=b

y=-4x+10

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Example Question #2 : Finding Equations Of Lines

Find a line through the point (2,3) perpendicular to the line 4x+y=7 .

Possible Answers:

y=4x+5/2

y=4x+7

y=1/4x+5/2

y=1/4x+7

7=-4x+5/2

Correct answer:

y=1/4x+5/2

Explanation:

To solve this problem you must first find the slope of the original equation by point it into y-mx+b form

4x+y=7

subtract x from both sides

y=-4x+7

Then you must find the reciprocal of the slope to get the slope of the perpendicular line

reciprocal is 1/4

Finally you must you point-slope form to solve

$y-y_{1}=m(x-x_{1})$

y-(3)=1/4(x-(2))

Multiply 1/4 through the parentheses

y-3=1/4x-1/2

add three to both sides

y=1/4x+5/2

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Example Question #1 : Circles

What is the equation of a circle with center at (4,-3) and a radius of ?

Possible Answers:

(x+4)^2+(y-3)^2=49

(x+4)^2+(y+3)^2=49

(x-4)^2+(y+3)^2=49

(x-4)^2+(y-3)^2=49

Correct answer:

(x-4)^2+(y+3)^2=49

Explanation:

Step 1: Recall the general equation for a circle (if the vertex is not at (0,0) :

(x-h)^2+(y-k)^2=r^2 , where center= (h,k)

Step 2: Recall the shift of the graph..

If the value of  is positive, it will be shown as a negative shift in the equation.
If the value of  is negative, it will be shown as a positive shift in the equation.
If the value of  is positive, it will be shown as a negative shift in the equation.
If the value of  is negative, it will be shown as a positive shift in the equation.

Step 3: Look at the center given in the problem and find the rule(s) in step 2 that will apply:

Center= (4,-3) , h=4 , k=-3

Step 4: Plug in h,k,r into the equation of a circle:

(x-4)^2+(y-(-3))^2=7^2
Simplify:

(x-4)^2+(y+3)^2=49

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Example Question #1 : Circles

What is the vertex of the equation of a circle: x^2+y^2=9

Possible Answers:

(-3,0)

(0,3)

(0,0)

(3,0)

Correct answer:

(0,0)

Explanation:

Step 1: There are no numbers next to and , so their is no movement of the vertex..

Step 2: Recall the vertex of a circle that does not move...

The vertex of this circle is (0,0) .

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Example Question #2 : Conic Sections

Using the information below, determine the equation of the hyperbola.

Foci: (-5,4) and (7,4)

Eccentricity: $\frac{3}{2}$

Possible Answers:

$\small \small \small \small \small \frac{(x-4)^{2}}{20} - \frac{(y-1)^{2}}{36}=1$

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

$\small \small \small \frac{(x-1)^{2}}{20} - \frac{(y-4)^{2}}{16}=1$

$\small \small \small \frac{(x-4)^{2}}{16} - \frac{(y-1)^{2}}{20}=1$

$\small \small \small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{36}=1$

Correct answer:

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

Explanation:

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

$\small \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}}=1$

Distance between foci =

Distance between vertices =

Eccentricity = c/a

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 12, we can set that equal to .

12=2c

$\small c= 6$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small c/a=3/2$

$\small \frac{c}{a}=\frac{3}{2}=\frac{6}{a}$

$\small 12=3a$

$\small a=4$

$\small \small a^{2}=16$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small 4^{2}+b^{2}=6^{2}$

$\small \small b^{2}=20$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 4, the center point will be on the same line. Hence, $\small k = 4$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 12, we can conclude that $\small h=1$

Center point: $\small \small (h, k)= (1,4)$

Thus, the equation of the hyperbola is:

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

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Example Question #11 : Conic Sections

Using the information below, determine the equation of the hyperbola.

Foci: (1, 8) and (9, 8)

Eccentricity:

Possible Answers:

$\small \small \small \small \small \small \small \frac{(x-8)^{2}}{4} - \frac{(y-5)^{2}}{12}=1$

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

$\small \small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{16}=1$

$\small \small \small \small \small \small \small \small \frac{(x-8)^{2}}{12} - \frac{(y-5)^{2}}{16}=1$

$\small \small \small \small \small \small \small \frac{(x-5)^{2}}{12} - \frac{(y-8)^{2}}{4}=1$

Correct answer:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

Explanation:

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

$\small \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}}=1$

Distance between foci =

Distance between vertices =

Eccentricity = c/a

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 8, we can set that equal to .

$\small 8=2c$

$\small \small c= 4$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small \small \small c/a=2$

$\small \small \frac{c}{a}=\frac{2}{1}=\frac{4}{a}$

$\small \small 4=2a$

$\small \small a=2$

$\small \small \small a^{2}=4$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small \small 2^{2}+b^{2}=4^{2}$

$\small \small \small b^{2}=12$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 8, the center point will be on the same line. Hence, $\small \small k = 8$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 8, we can conclude that $\small \small h=5$

Center point: $\small \small \small (h, k)= (5, 8 )$

Thus, the equation of the hyperbola is:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

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Example Question #2 : Solve A System Of Quadratic Equations

Find the coordinate of intersection, if possible: y=x^2+4 and y=-x^2+4 .

Possible Answers:

$(0,\pm4)$

(-4,0)

$\textup{There is no intersection.}$

(4,0)

(0,4)

Correct answer:

(0,4)

Explanation:

To solve for x and y, set both equations equal to each other and solve for x.

-x^2+4=x^2+4

2x^2 =0

x=0

Substitute x=0 into either parabola.

y=0^2+4 = 4

The coordinate of intersection is (0,4) .

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GRE Subject Test: Math : GRE Subject Test: Math

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Finding Equations Of Lines

Example Question #1 : Finding Equations Of Lines

Example Question #2 : Finding Equations Of Lines

Example Question #1 : Finding Equations Of Lines

Example Question #2 : Finding Equations Of Lines

Example Question #1 : Circles

Example Question #1 : Circles

Example Question #2 : Conic Sections

Example Question #11 : Conic Sections

Example Question #2 : Solve A System Of Quadratic Equations

All GRE Subject Test: Math Resources

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