Calculating x or y intercept - GMAT Quantitative Reasoning

Card 0 of 20

Question

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

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

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

Compare your answer with the correct one above

Question

What is the -intercept of the line ?

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

Compare your answer with the correct one above

Question

What is the -intercept of the line ?

Answer

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

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

The -intercept is .

Compare your answer with the correct one above

Question

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

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 $\left (0, 6\frac{3}{5} \right )$

Compare your answer with the correct one above

Question

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

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:

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

Compare your answer with the correct one above

Question

What is the -intercept of

Answer

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

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

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

Compare your answer with the correct one above

Question

What is -intercept for

Answer

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

Compare your answer with the correct one above

Question

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

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:

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

A line includes and . Give its -intercept.

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.

Compare your answer with the correct one above

Question

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

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

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

Compare your answer with the correct one above

Question

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

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)

Compare your answer with the correct one above

Question

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

$4 x+ \bigcirc y = 18$

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.

Compare your answer with the correct one above

Question

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

$\bigcirc x+ 5 y = 18$

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.

Compare your answer with the correct one above

Question

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

$4x+5y = \bigcirc$

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.

Compare your answer with the correct one above

Question

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

$4 x+ \bigcirc y = 18$

Answer

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

Substitute 0 for and 5 for ; the equation becomes

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

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

Compare your answer with the correct one above

Question

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

$4x+5y = \bigcirc$

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.

Compare your answer with the correct one above

Question

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

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

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.

Compare your answer with the correct one above

Question

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

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

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.

Compare your answer with the correct one above

Question

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

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

Compare your answer with the correct one above

Question

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

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.

Compare your answer with the correct one above

Tap the card to reveal the answer