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

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What is the -intercept of the graph of the function

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

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

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

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

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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 .

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

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

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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.

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

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

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

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

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

where

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 .

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

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

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

where

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 )$ .

Give the -intercept of the line that passes through points and .

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

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

setting :

$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

; the line becomes

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

or

$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 )$ .

Give the -intercept of the line that passes through points and .

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

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

setting :

$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

; the line becomes

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

or

$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 )$ .

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

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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 parallel to the second, has the same slope.

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

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

with

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

the equation becomes

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

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

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

$- 2 = $\frac{3}{4}$ x - $\frac{15}{4}$$

$- 2 + $\frac{15}{4}$= $\frac{3}{4}$ x - $\frac{15}{4}$ + $\frac{15}{4}$$

$\frac{3}{4}$ x = $\frac{7}{4}$

$\frac{4}$ {3}\cdot $\frac{3}{4}$ x =\frac{4}$ {3}\cdot $\frac{7}{4}$

$x = $\frac{7}{3}$ = 2$\frac{1}{3}$$

The -intercept is the point $\left (2 $\frac{1}{3}$, 0 \right )$ .

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

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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 .

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

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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 )$ .

Find the y-intercept:

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Rewrite the equation in slope-intercept form, .

The y-intercept is , which is .

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

(a)

(b)

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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$

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

Similarly, substituting 3 for :

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

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

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

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

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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 .

What is the -intercept of the graph of the function

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

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

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

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

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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 .

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

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

Tap to see back →

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.

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

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

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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, $\frac{3}{8}\right )$ .