SSAT Upper Level Quantitative - How to find x or y intercept

Define . The graphs of and a second function, , intersect at their common -intercept. Which of the following could be the definition of ?

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 , which can be found as follows:

Substituting the definition, we get

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

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

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

The -intercept of the graph of 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 $g \left ( - \frac{7}{2} \right ) = 0$ .

can be eliminated immediately as a choice since it cannot take the value 0.

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

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

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

The correct choice is .

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

(a)

(b)

Neither (a) nor (b)

Both (a) and (b)

(b), but not (a)

(a), but not (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 .

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

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

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

, so is not an -intercept.

Similarly, substituting 3 for :

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

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

, so is not an -intercept.

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

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 , 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 .

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

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

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

The line has no -intercept.

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 :

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

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

The -intercept is the point .

What is the -intercept of the graph of the function

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

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

$\left (0, 3 \right )$

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

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

The graph has no -intercept.

Explanation

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

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

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

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

Give the -intercept, if there is one, of the graph of the equation

$y = \frac{1}{x^{2}+ x} + 3$

The graph has no -intercept.

$\left ( 0, \frac{3}{2}\right )$

Explanation

The -intercept is the point at which the graph crosses the -axis; at this point, the -coordinate is 0, so substitute for in the equation:

$y = \frac{1}{x^{2}+ x} + 3$

$y = \frac{1}{0^{2}+0} + 3$

$y = \frac{1}{0} + 3$

However, since this expression has 0 in a denominator, it is of undefined value. This means that there is no value of paired with -coordinate 0, and, subsequently, the graph of the equation has no -intercept.

What is the -intercept of the graph of the function $f \left ( x\right ) = 2x^{2} - 7x + 5$ ?

$\left (0,5 \right )$

$\left (0,-7 \right )$

$\left (0,2 \right )$

$\left (0,0 \right )$

$\left ( 0, -3 \frac{1}{2}\right )$

Explanation

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

$f \left ( x\right ) = 2x^{2} - 7x + 5$

$f \left (0\right ) = 2 \cdot 0^{2} - 7 \cdot 0 + 5$

$f \left (0\right ) = 0- 0 + 5 = 5$

The -intercept is $\left (0,5 \right )$ .

Give the -intercept, if there is one, of the graph of the equation

$y = \frac{3}{\left | 3x-8\right |}$

$\left ( 0, \frac{3}{8}\right )$

The graph has no -intercept.

$\left ( 0, -\frac{3}{8}\right )$

$\left ( 0, \frac{3}{5}\right )$

$\left ( 0,- \frac{3}{5}\right )$

Explanation

The -intercept is the point at which the graph crosses the -axis; at this point, the -coordinate is 0, so substitute for in the equation:

$y = \frac{3}{\left | 3x-8\right |}$

$y = \frac{3}{\left | 3 \cdot 0 -8\right |}$

$y = \frac{3}{\left | 0 -8\right |}$

$y = \frac{3}{\left | -8\right |}$

$y = \frac{3}{ 8}$

The -intercept is $\left ( 0, \frac{3}{8}\right )$ .

Give the -intercept, if there is one, of the graph of the equation

$y = 3(x-2)^{2} + 4 (x-7)$ .

The graph does not have a -intercept.

Explanation

The -intercept is the point at which the graph crosses the -axis; at this point, the -coordinate is 0, so substitute for in the equation:

$y = 3(x-2)^{2} + 4 (x-7)$

$y = 3(0-2)^{2} + 4 (0-7)$

$y = 3( -2)^{2} + 4 ( -7)$

$y = 3 \cdot 4 + 4 ( -7)$

The -intercept is the point .

A line passes through and is perpendicular to the line of the equation . Give the -intercept of this line.

$\left (6 \frac{1}{2}, 0 \right )$

$\left (2 \frac{1}{3}, 0 \right )$

$\left ( -1\frac{3}{4}, 0\right )$

$\left ( 8\frac{2}{3}, 0\right )$

The line has no -intercept.

Explanation

First, find the slope of the second line by solving for as follows:

$\frac{4y }{4}=\frac{ 3x-17 }{4}$

$y = \frac{3}{4}x- \frac{17}{4}$

The equation is now in the slope-intercept form ; the slope of the second line is the -coefficient $m= \frac{3}{4}$ .

The first line, being perpendicular to the second, has as its slope the opposite of the reciprocal of $\frac{3}{4}$ , which is $m=- \frac{4}{3}$ .

Therefore, we are looking for a line through with slope $m=- \frac{4}{3}$ . Using point-slope form

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

with

$x _{1 } =5, y _{1 } = 2,m=- \frac{4}{3}$ ,

the equation becomes

$y - 2 =- \frac{4}{3}(x -5)$ .

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

$0 - 2 =- \frac{4}{3}(x -5)$

$- 2 =- \frac{4}{3}x+ \frac{20}{3}$

$- 2- \frac{20}{3} =- \frac{4}{3}x+ \frac{20}{3} - \frac{20}{3}$

$- \frac{26}{3} =- \frac{4}{3}x$

$- \frac{3} {4}\cdot\left (- \frac{26}{3} \right )=- \frac{3} {4}\cdot \left (- \frac{4}{3}x \right )$

$x= \frac{26}{4} = \frac{13}{2} = 6\frac{1}{2}$

The -intercept is the point $\left (6 \frac{1}{2}, 0 \right )$ .

How to find x or y intercept

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