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SSAT Upper Level Math : Coordinate Geometry

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Example Questions

Example Question #1 : X And Y Intercept

Give the -intercept of the line that passes through points (-3, 4) and (2, -3) .

Possible Answers:

The line has no -intercept.

$\left (0, - \frac{1}{7} \right )$

$\left (0, - 5\frac{6}{7} \right )$

$\left (0, - \frac{1}{5} \right )$

$\left (0, - 8 \frac{1}{5} \right )$

Correct answer:

$\left (0, - \frac{1}{5} \right )$

Explanation:

First, find the slope of the line, using the slope formula

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

setting $x_{1} = -3, y_{1} = 4, x_{2} = 2, y_{2} = -3$ :

$m = \frac{-3 -4}{2-(-3)} = \frac{-7}{5} =- \frac{7}{5}$

By the point-slope formula, this line has the equation

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

where

$x_{1} = -3, y_{1} = 4, m = - \frac{7}{5}$ ; the line becomes

$y - 4 = - \frac{7}{5}(x -(-3))$

$y - 4 = - \frac{7}{5}(x +3)$

To find the -intercept, substitute 0 for and solve for :

$y - 4 = - \frac{7}{5}(0+3)$

$y - 4 = - \frac{7}{5}(3)$

$y - 4 = - \frac{21}{5}$

$y - 4+ 4 = - \frac{21}{5} + 4$

$y = - \frac{1}{5}$

The -intercept is $\left (0, - \frac{1}{5} \right )$ .

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Example Question #8 : How To Find X Or Y Intercept

Give the -intercept of the line that passes through points (-3, 4) and (2, -3) .

Possible Answers:

$\left (- \frac{1}{7} , 0 \right )$

$\left ( - \frac{1}{5},0 \right )$

$\left ( - 8 \frac{1}{5},0 \right )$

$\left (- 5\frac{6}{7} , 0 \right )$

$\text{The line has no } x\text{-intercept.}$

Correct answer:

$\left (- \frac{1}{7} , 0 \right )$

Explanation:

First, find the slope of the line, using the slope formula

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

setting $x_{1} = -3, y_{1} = 4, x_{2} = 2, y_{2} = -3$ :

$m = \frac{-3 -4}{2-(-3)} = \frac{-7}{5} =- \frac{7}{5}$

By the point-slope formula, this line has the equation

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

where

$x_{1} = -3, y_{1} = 4, m = - \frac{7}{5}$ ; the line becomes

$y - 4 = - \frac{7}{5}(x -(-3))$

$y - 4 = - \frac{7}{5}(x +3)$

To find the -intercept, substitute 0 for and solve for :

$0 - 4 = - \frac{7}{5}(x +3)$

$- 4 = - \frac{7}{5} x - \frac{21}{5}$

$- 4+ \frac{21}{5} = - \frac{7}{5} x - \frac{21}{5}+ \frac{21}{5}$

$\frac{1}{5} = - \frac{7}{5} x$

$- \frac{5}{7} \cdot \frac{1}{5} = - \frac{5}{7}\left ( - \frac{7}{5} x \right )$

$- \frac{1}{7} =x$

The -intercept is $\left (- \frac{1}{7} , 0 \right )$ .

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Example Question #101 : Expressions & Equations

Give the -intercept of the line with slope $\frac{2}{5}$ that passes through point (5, 9) .

Possible Answers:

(7, 0)

$\left ( -27 \frac{1}{2}, 0 \right )$

(-11, 0)

$\left ( -17 \frac{1}{2}, 0 \right )$

$\text{The line has no } x \text{-intercept.}$

Correct answer:

$\left ( -17 \frac{1}{2}, 0 \right )$

Explanation:

By the point-slope formula, this line has the equation

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

where

$x_{1} = 5, y_{1} = 9 ,m = \frac{2}{5}$

By substitution, the equation becomes

$y -9 = \frac{2}{5}(x -5)$

To find the -intercept, substitute 0 for and solve for :

$0-9 = \frac{2}{5}(x -5)$

$-9 = \frac{2}{5}x -2$

$-9 +2= \frac{2}{5}x -2+2$

$-7 = \frac{2}{5}x$

$\frac{5}{2} \cdot \left (-7 \right ) = \frac{5}{2} \cdot \frac{2}{5}x$

$x = -\frac{35}{2} = -17 \frac{1}{2}$

The -intercept is $\left ( -17 \frac{1}{2}, 0 \right )$ .

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Example Question #501 : Ssat Upper Level Quantitative (Math)

Find the y-intercept: -y-8=-6x+2

Possible Answers:

Correct answer:

Explanation:

Rewrite the equation in slope-intercept form, y=mx+b .

-y-8=-6x+2

-y=-6x+10

y=6x-10

The y-intercept is , which is .

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Example Question #11 : X And Y Intercept

What is the -intercept of the graph of the function

$f(x) = x^{2} + 3x - 10?$

Possible Answers:

$\left (0, 12 \frac{1}{4} \right )$

$\left (0, 3 \frac{1}{3} \right )$

The graph has no -intercept.

$\left (0, -10 \right )$

$\left (0, 3 \right )$

Correct answer:

$\left (0, -10 \right )$

Explanation:

The -intercept of the graph of a function is the point at which it intersects the -axis - that is, at which x = 0 . This point is $\left (0, f(0) \right )$ , so evaluate f(0) :

$f(x) = x^{2} + 3x - 10$

$f( 0) = 0^{2} + 3 \cdot 0 - 10 = 0 + 0 - 10 = -10$

The -intercept is $\left (0, -10 \right )$ .

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Example Question #222 : Coordinate Geometry

Define a function $f(x) = x ^{2}+ 3x + 2$ . Which of the following is the -intercept of the graph of ?

Possible Answers:

(0, -2)

(0,2)

(0, 3)

(0, -1)

Correct answer:

(0,2)

Explanation:

The -intercept of the graph of a function has 0 as its -coordinate, since it is defined to be the point at which it crosses the -axis. Its -coordinate is f(0) , which can be found using substitution, as follows:

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

$f(0) = 0 ^{2}+ 3 \cdot 0 + 2 = 0 + 0 + 2 = 2$

The correct choice is (0,2) .

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Example Question #221 : Coordinate Geometry

Define a function $f(x) = x ^{2}+ 3x + 2$ . Which of the following is an -intercept of the graph of ?

(a) (2, 0)

(b) (3, 0)

Possible Answers:

Both (a) and (b)

(a), but not (b)

(b), but not (a)

Neither (a) nor (b)

Correct answer:

Neither (a) nor (b)

Explanation:

An -intercept of the graph of a function has 0 as its -coordinate, since it is defined to be a point at which it crosses the -axis. Its -coordinate is a value of for which f(x) = 0 .

We can most easily determine whether (2, 0) is a point on the graph of f(x) by proving or disproving that f(2) = 0 , which we can do by substituting 2 for :

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

$f(2) = 2^{2}+ 3 \cdot 2 + 2$

f(2) = 4+ 6 + 2

f(2) = 1 2

$f(2) \ne 0$ , so (2, 0) is not an -intercept.

Similarly, substituting 3 for :

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

$f(3) = 3^{2}+ 3 \cdot 3 + 2$

f(3) = 9+ 9 + 2

f(3) = 1 2

$f(3) \ne 0$ , so (3, 0) is not an -intercept.

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Example Question #281 : Geometry

Define f(x) = 2x+ 7 . The graphs of f(x) and a second function, g(x) , intersect at their common -intercept. Which of the following could be the definition of g(x) ?

Possible Answers:

g(x) = 2x+ 14

g(x) = 4x+ 14

g(x) = 4x+ 7

g(x) = 7

Correct answer:

g(x) = 4x+ 14

Explanation:

An -intercept of the graph of a function has 0 as its -coordinate, since it is defined to be a point at which it crosses the -axis. Its -coordinate is a value of for which f(x) = 0 , which can be found as follows:

Substituting the definition, we get

2x+ 7 = 0

Solving for by subtracting 7 from both sides, then dividing both sides by 2:

2x+ 7 -7 = 0 -7

2x = -7

$2x \div 2 = -7 \div 2$

$x= - \frac{7}{2}$

The -intercept of the graph of f(x) is the point $\left ( - \frac{7}{2}, 0 \right )$ .

To determine which of the four choices is correct, substitute $- \frac{7}{2}$ for and determine for which definition of it holds that $g \left ( - \frac{7}{2} \right ) = 0$ .

g(x) = 7 can be eliminated immediately as a choice since it cannot take the value 0.

g(x) = 2x+ 14

$g\left ( - \frac{7}{2} \right ) = 2\left ( - \frac{7}{2} \right ) + 14 = -7 + 14 = 7$

g(x) = 4x+ 7 :

$g \left ( - \frac{7}{2} \right ) = 4 \left ( - \frac{7}{2} \right ) + 7 = -14 + 7 = -7$

g(x) = 4x+ 14

$g \left ( - \frac{7}{2} \right ) = 4 \left ( - \frac{7}{2} \right ) + 14 = -14 + 14 =0$

The correct choice is g(x) = 4x+ 14 .

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Example Question #1 : How To Find The Equation Of A Curve

If the -intercept of the line is and the slope is , which of the following equations best satisfies this condition?

Possible Answers:

x-2y=-6

-2x-y=6

y=x+6

y=x-3

y=x-6

Correct answer:

y=x-6

Explanation:

Write the slope-intercept form.

y=mx+b

The point given the x-intercept of 6 is (6,0) .

Substitute the point and the slope into the equation and solve for the y-intercept.

0=(1)(6)+b

b=-6

Substitute the y-intercept back to the slope-intercept form to get your equation.

y=x-6

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Example Question #2 : How To Find The Equation Of A Curve

A vertical parabola on the coordinate plane has vertex (-4, 10) and -intercept (0, -2) .

Give its equation.

Possible Answers:

$y = \frac{3}{4} x^{2} + 6x+22$

$y = -\frac{3}{4} x^{2}- 6x-2$

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

Insufficient information is given to determine the equation.

$y = \frac{1}{2} x^{2}+4x+18$

Correct answer:

$y = -\frac{3}{4} x^{2}- 6x-2$

Explanation:

The equation of a vertical parabola, in vertex form, is

$y = a(x-h)^{2}+k$ ,

where (h,k) is the vertex. Set h = -4,k=10 :

$y = a[x-(-4)]^{2}+10$

$y = a(x+4)^{2}+10$

To find , use the -intercept, setting x=0,y= -2 :

$-2 = a(0+4)^{2}+10$

$-2 = a\cdot 4^{2}+10$

-2 = 16a +10

-12 = 16a

$a =- \frac{3}{4}$

The equation, in vertex form, is $y =- \frac{3}{4}(x+4)^{2}+10$ ; in standard form:

$y = -\frac{3}{4}(x+4)^{2}+10$

$y =- \frac{3}{4}(x^{2} + 8x+16)+10$

$y = -\frac{3}{4} x^{2}- 6x-12+10$

$y = -\frac{3}{4} x^{2}- 6x-2$

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SSAT Upper Level Math : Coordinate Geometry

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #1 : X And Y Intercept

Example Question #8 : How To Find X Or Y Intercept

Example Question #101 : Expressions & Equations

Example Question #501 : Ssat Upper Level Quantitative (Math)

Example Question #11 : X And Y Intercept

Example Question #222 : Coordinate Geometry

Example Question #221 : Coordinate Geometry

Example Question #281 : Geometry

Example Question #1 : How To Find The Equation Of A Curve

Example Question #2 : How To Find The Equation Of A Curve

All SSAT Upper Level Math Resources

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