Calculating x or y intercept

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GMAT Quantitative › Calculating x or y intercept

Questions 1 - 10

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

Explanation

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

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

$9 \frac{3}{8}$

$18\frac{3}{4}$

$7\frac{1}{2}$

Explanation

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 :

Triangle_2

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:

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

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

$4 x+ \bigcirc y = 18$

$\frac{18 }{5}$

$\frac{4}{5}$

$\frac{5}{4}$

$\frac{5}{18}$

Explanation

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

Substitute 0 for and 5 for ; the equation becomes

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

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

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

$4x+5y = \bigcirc$

$\frac{6}{5}$

$\frac{3}{2}$

The graph cannot have as its -intercept regardless of the value written in the circle.

Explanation

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.

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

$4x+5y = \bigcirc$

$\frac{7}{5}$

$\frac{7}{4}$

Explanation

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.

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

$\bigcirc x+ 5 y = 18$

$-\frac{5}{6}$

The graph cannot have as its -intercept regardless of the value written in the circle.

$-\frac{6}{5}$

Explanation

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.

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

$4 x+ \bigcirc y = 18$

The graph cannot have as its -intercept regardless of the value written in the circle.

$-\frac{18}{5}$

$-\frac{9}{10}$

$-\frac{5}{4}$

Explanation

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.

A line includes and . Give its -intercept.

The line has no -intercept.

Explanation

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.

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

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

The graph will have at least one -intercept regardless of the value written in the circle.

Explanation

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.

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

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

None of the other choices is correct.

Explanation

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