Calculating x or y intercept - GMAT Quantitative

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Question

Give the area of the region on the coordinate plane bounded by the -axis, the -axis, and the graph of the equation .

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Answer

This can best be solved using a diagram and noting the intercepts of the line of the equation , which are calculated by substituting 0 for and separately and solving for the other variable.

-intercept:

$3x + 4 \cdot 0 = 15$

-intercept:

$3 \cdot 0 + 4y = 15$

$y = 3 \frac{3}{4}$

Now, we can make and examine the diagram below - the red line is the graph of the equation :

The pink triangle is the one whose area we want; it is a right triangle whose legs, which can serve as base and height, are of length $5, 3\frac{3}{4}$ . We can compute its area:

$\frac{1}{2} \cdot 5 \cdot 3\frac{3}{4}= \frac{1}{2} \cdot \frac{5}{1} \cdot \frac{15}{4} = \frac{75}{8} = 9 \frac{3}{8}$

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Question

Give the -intercept(s) of the graph of the equation

$y = 3x^{2} -2x - 16$

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Answer

Set

$y = 3x^{2} -2x - 16$

$3x^{2} -2x - 16 = 0$

Using the -method, we look to split the middle term of the quadratic expression into two terms. We are looking for two integers whose sum is and whose product is $3 \left ( -16 \right ) = -48$ ; these numbers are .

$3x^{2} -8x + 6x - 16 = 0$

$\left (3x^{2} -8x \right )+ \left ( 6x - 16 \right ) = 0$

$x \left (3x -8 \right )+ 2 \left (3 x - 8 \right ) = 0$

$\left ( x + 2 \right )\left (3x -8 \right ) = 0$

Set each linear binomial to 0 and solve:

or

$3x\div 3 = 8 \div 3$

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

There are two -intercepts - $(-2,0), \left ( \frac{8}{3}, 0 \right )$

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Question

What is the -intercept of the line ?

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Answer

Substitute 0 for and solve for :

$A \cdot 0 + 2Ay = 3A$

$2Ay \div 2A = 3A\div 2A$

$y = \frac{3A}{2A} = \frac{3}{2}$

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

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Question

What is the -intercept of the line ?

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Answer

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

$x+ 2A \cdot 0= 4A$

The -intercept is .

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Question

What is the -intercept of a line that includes points and ?

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Answer

The slope of the line is

$m = \frac{y _{2}- y _{1}}{x _{2}- x _{1}}=\frac{1- 5}{7- 2} = \frac{-4}{5} = - \frac{4}{5}$

Use the point slope form to find the equation of the line.

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

$y - 5= - \frac{4}{5} \left ( x - 2\right )$

Now substitute and solve for .

$y - 5= - \frac{4}{5} \left ( 0 - 2\right )$

$y - 5= \frac{8}{5}$

$y = \frac{8}{5} + 5 = \frac{8}{5} + \frac{25}{5} = \frac{33}{5} = 6\frac{3}{5}$

The -intercept is $\left (0, 6\frac{3}{5} \right )$

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Question

What is the -intercept of

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Answer

To solve for the -intercept, you have to set to zero and solve for :

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

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

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Question

What is -intercept for

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Answer

To solve for the -intercept, you have to set to zero and solve for :

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Question

A line with slope $-\frac{4}{5}$ includes point . What is the -intercept of this line in terms of ?

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Answer

For some real number , the -intercept of the line will be some point . We can set up the slope equation and solve for as follows:

$\frac{y_2-y_1}{x_2-x_1} = m$

$\frac{b-N}{0-6} =-\frac{4}{5}$

$\frac{b-N}{-6} =-\frac{4}{5}$

$\frac{b-N}{-6} \cdot (-6)=-\frac{4}{5}\cdot (-6)$

$b-N = \frac{24}{5}$

$b-N+N =N+ \frac{24}{5}$

$b =N+ \frac{24}{5}$

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Question

A line includes and . Give its -intercept.

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Answer

The two points have the same coordinate, which is 5; the line is therefore vertical. This makes the line parallel to the -axis, meaning that it does not intersect it. Therefore, the line has no -intercept.

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Question

Give the -intercept(s) of the graph of the equation

$y = 3x^{2} -2x - 16$

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Answer

Substitute 0 for :

$y = 3x^{2} -2x - 16$

$y = 3 \cdot 0^{2} -2 \cdot 0 - 16 = 0 - 0 -16 = -16$

The -intercept is

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Question

What are the and intercepts of the function $y= \ln(x+3)$ ?

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Answer

The correct answer is

y-intercept at $(0,\ln(3))$

x-intercept at

To find the y-intercept, we plug in for and solve for

So we have $y = \ln(3+0) = \ln(3)$ . This is as simplified as we can get.

To find the x-intercept, we plug in for and solve for

So we have $0 = \ln(x+3)$

$e^0 = e^{\ln(x+3)}$ (Exponentiate both sides)

( is 1, and cancel the and ln on the right side)

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Question

Fill in the circle with a number so that the graph of the resulting equation has -intercept :

$4 x+ \bigcirc y = 18$

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Answer

Let be the number in the circle. The equation can be written as

Substitute 0 for ; the resulting equation is

$4 x+N \cdot 0 = 18$

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

The -intercept is $\left ( \frac{9}{2}, 0 \right )$ regardless of what number is written in the circle.

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Question

Fill in the circle with a number so that the graph of the resulting equation has -intercept :

$\bigcirc x+ 5 y = 18$

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Answer

Let be the number in the circle. The equation can be written as

Substitute 0 for and for ; the resulting equation is

$N (-3)+ 5 \cdot 0= 18$

is the correct choice.

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Question

Fill in the circle with a number so that the graph of the resulting equation has -intercept :

$4x+5y = \bigcirc$

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Answer

Let be the number in the circle. The equation can be written as

Substitute 0 for and 6 for ; the resulting equation is

$4 \cdot 6+5 \cdot 0 = N$

24 is the correct choice.

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Question

Fill in the circle with a number so that the graph of the resulting equation has -intercept :

$4 x+ \bigcirc y = 18$

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Answer

Let be the number in the circle. The equation can be written as

Substitute 0 for and 5 for ; the equation becomes

$4 \cdot 0+N \cdot 5 = 18$

$N = \frac{18 }{5}$

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Question

Fill in the circle with a number so that the graph of the resulting equation has -intercept :

$4x+5y = \bigcirc$

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Answer

Let be the number in the circle. The equation can be written as

Substitute 7 for and 0 for ; the resulting equation is

$4 \cdot 0+5 \cdot 7= N$

35 is the correct choice.

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Question

Fill in the circle so that the graph of the resulting equation has no -intercepts:

$y = 4x^{2}+ 6 x + \bigcirc$

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Answer

Let be the number in the circle. Then the equation can be rewritten as

$y = 4x^{2}+ 6 x + C$

Substitute 0 for and the equation becomes

$4x^{2}+ 6 x + C = 0$

Equivalently, we are seeking a value of for which this equation has no real solutions. This happens in a quadratic equation $Ax^{2}+ Bx+ C = 0$ if and only if

$B^{2}-4AC < 0$

Replacing with 4 and with 6, this becomes

$6^{2}-4 \cdot 4 \cdot C < 0$

$C > \frac{36}{16} = \frac{9}{4}$

Therefore, must be greater than $\frac{9}{4}$ . The only choice fitting this requirement is 4, so this is correct.

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Question

Fill in the circle so that the graph of the resulting equation has exactly one -intercept:

$y = 4x^{2}+ \bigcirc x + 8$

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Answer

Let be the number in the circle. Then the equation can be rewritten as

$y = 4x^{2}+ Bx + 8$

Substitute 0 for and the equation becomes

$4x^{2}+ Bx + 8 = 0$

Equivalently, we are seeking a value of for which this equation has exactly one solution. This happens in a quadratic equation $Ax^{2}+ Bx+ C = 0$ if and only if

$B^{2}-4AC = 0$

Replacing with 4 and with 8, this becomes

$B^{2}-4 \cdot 4 \cdot 8 = 0$

$B^{2}-128= 0$

$B^{2}=128$

$B= \pm \sqrt{128} = \pm \sqrt{64} \cdot \sqrt{2} = \pm 8 \sqrt{2}$

Therefore, either $B=- 8 \sqrt{2}$ or $B= 8 \sqrt{2}$ .

Neither is a choice.

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Question

Find the $y\textup{-intercept}$ for the following equation:

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Answer

To find the $y\textup{-intercept}$ , you must put the equation into slope intercept form:

where is the intercept.

Thus,

$y=\frac{3}2{x-3}$

Therefore, your $y\textup{-intercept}$ is

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Question

Find where g(x) crosses the y-axis.

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Answer

Find where g(x) crosses the y-axis.

A function will cross the y-axis wherever x is equal to 0. This may be easier to see on a graph, but it can be thought of intuitively as well. If x is 0, then we are neither left nor right of the y-axis. This means we must be on the y-axis.

So, find g(0)

So our answer is 945.

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