x and y intercept - GMAT Quantitative

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Question

Suppose the points and are plotted to connect a line. What are the -intercept and -intercept, respectively?

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Answer

First, given the two points, find the equation of the line using the slope formula and the y-intercept equation.

Slope:

$\frac{y_2-y_1}{x_2-x_1} = \frac{3-(-2)}{1-(-3)}=\frac{5}{4}$

Write the slope-intercept formula.

Substitute a given point and the slope into the equation to find the y-intercept.

$3=\left(\frac{5}{4}\right)(1)+b$

$\frac{12}{4}=\frac{5}{4}+b$

$b=\frac{7}{4}$

The y-intercept is: $b=\frac{7}{4}$ .

Substitiute the slope and the y-intercept into the slope-intercept form.

$y=\frac{5}{4}x+\frac{7}{4}$

To find the x-intercept, substitute and solve for x.

$0=\frac{5}{4}x+\frac{7}{4}$

$-\frac{7}{4}=\frac{5}{4}x$

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

$x=-\frac{7}{5}$

The x-intercept is: $x=-\frac{7}{5}$

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Question

Suppose the curve of a function is parabolic. The -intercept is and the vertex is the -intercept at . What is a possible equation of the parabola, if it exists?

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Answer

Write the standard form of the parabola.

Given the point , the y-intercept is -4, which indicates that . This is also the vertex, so the vertex formula can allow writing an expression in terms of variables and .

Write the vertex formula and substitute the known vertex given point .

$x=-\frac{b}{2a}$

$0=-\frac{b}{2a}$

Using the values of , , and the other given point , substitute these values to the standard form and solve for .

Substitute the values of ,, and into the standard form of the parabola.

The correct answer is:

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Question

If the -intercept and the slope are , what's the equation of the line in standard form?

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Answer

Write the slope intercept formula.

Convert the given x-intercept to a known point, which is .

Substitute the given slope and the point to solve for the y-intercept.

Substitute the slope and y-intercept into the slope-intercept formula.

Add 1 on both sides of the equation, and subtract on both sides of the equation to find the equation in standard form.

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Question

Which of the following functions has as its graph a curve with , and as its only two -intercepts?

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Answer

By the Fundamental Theorem of Algebra, a polynomial equation of degree 3 must have three solutions, or roots, but one root can be a double root or triple root. Since the polynomial here has two roots, and 4, one of these must be a double root. Since the leading term is $x^{2}$ , the equation must be

or

We rewrite both.

$(x+4)(x^{2}-4^{2}) = 0$

$(x+4)(x^{2}-16) = 0$

$x \cdot x^{2}- x \cdot 16 + 4 \cdot x^{2}- 4 \cdot 16 = 0$

$x^{3}-16 x + 4 x^{2}- 64 = 0$

$x^{3}+ 4 x^{2}-16 x - 64 = 0$

$(x^{2}-16) (x-4)= 0$

$x \cdot x^{2}- 4 \cdot x^{2}- x \cdot 16 + 4 \cdot 16 = 0$

$x^{3}- 4 x^{2}-16 x + 64 = 0$

The correct response can be $f(x) = x^{3}+ 4 x^{2}-16 x - 64$ or $f(x) = x^{3}- 4 x^{2}-16 x + 64$ . The first is not among the choices, so the last is the correct choice.

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Question

Which of the following functions does not have as its graph a curve with as an -intercept?

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Answer

We can evaluate in each of the definitions of in the five choices. If , is an -intercept.

$f(x) = x^{3}+ 5x^{2} -25 x-125$

$f(-5) = (-5)^{3}+ 5 \cdot (-5)^{2} -25 (-5)-125$

$= (-5)^{3}+ 5 \cdot 25 - (-125)-125$

$f(x) = x^{3}- 5x^{2} -25 x+125$

$f(-5) = (-5)^{3}- 5 \cdot (-5)^{2} -25 (-5)+125$

$= (-5)^{3}+ 5 \cdot 25 - (-125)-125$

$f(x) = x^{3} +125$

$f(-5) = (-5) ^{3} +125$

$f(x) = x^{3}+1 5x^{2} + 75 x+125$

$f(-5) = (-5)^{3}+1 5\cdot (-5)^{2} + 75 (-5)+125$

$f(-5) = -125+1 5\cdot 25 + (-375)+125$

$f(x) = x^{3}-1 5x^{2} + 75 x-125$

$f(-5) = (-5)^{3}-1 5\cdot (-5)^{2} + 75 (-5)-125$

$f(-5) = -125-1 5\cdot 25 + (-375)-125$

$f(x) = x^{3}-1 5x^{2} + 75 x-125$ does not have as an -intercept, so it is the correct choice.

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Question

Only one of the following equations has a graph with an -intercept between $\left ( \frac{1}{2},0 \right )$ and . Which one?

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Answer

The Intermediate Value Theorem states that if is a continuous function, as all five of the polynomial functions in the given choices are, and $f\left ( \frac{1}{2} \right )$ and are of different sign, then the graph of has an -intercept on the interval $\left ( \frac{1}{2}, 1 \right )$ .

We evaluate $f\left ( \frac{1}{2} \right )$ and for each of the five choices to find the one for which the two have different sign.

$f(x) = x^{4}-x^{3}+ x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+ \frac{1}{2} -3 = \frac{1}{16} - \frac{1}{8}+ \frac{1}{2}-3=-2\frac{9}{16}$

$f(1) = 1^{4}-1^{3}+ 1-3 = 1-1+1-3 = -2$

$f\left ( \frac{1}{2} \right )$ and are both negative.

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

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+2 \left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+1-3=-2\frac{1}{16}$

$f(1) = 1^{4}-1^{3}+ 2 \cdot 1-3 = 1-1+2-3 = -1$

$f\left ( \frac{1}{2} \right )$ and are both negative.

$f(x) = x^{4}-x^{3}+ 4 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+4 \left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+2-3=-1\frac{1}{16}$

$f(1) = 1^{4}-1^{3}+ 4 \cdot 1-3 = 1-1+4-3 = 1$

$f\left ( \frac{1}{2} \right )$ and are of different sign.

$f(x) = x^{4}-x^{3}+ 8 x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+8\left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+4-3= \frac{15}{16}$

$f(1) = 1^{4}-1^{3}+ 8 \cdot 1-3 = 1-1+8-3 = 5$

$f\left ( \frac{1}{2} \right )$ and are both positive.

$f(x) = x^{4}-x^{3}+ 16x-3$

$f\left ( \frac{1}{2} \right )= \left ( \frac{1}{2} \right )^{4}-\left ( \frac{1}{2} \right )^{3}+16\left (\frac{1}{2} \right )-3 = \frac{1}{16} - \frac{1}{8}+8-3=4 \frac{15}{16}$

$f(1) = 1^{4}-1^{3}+ 16 \cdot 1-3 = 1-1+16-3 = 13$

$f\left ( \frac{1}{2} \right )$ and are both positive.

$f(x) = x^{4}-x^{3}+ 4 x-3$ is the function in which $f\left ( \frac{1}{2} \right )$ and are of different sign, so it is represented by a graph with an -intercept between $\left ( \frac{1}{2},0 \right )$ and . This is the correct choice.

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Question

Between which two points is an -intercept of the graph of the function

$f(x) = x^{4}+3x^{2}-17x-115$

located?

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Answer

As a polynomial function, has a continuous graph. By the Intermediate Value Theorem, if and are of different sign, then for some $c \in (a,b)$ - that is, the graph of has an -intercept between and . Evaluate for all and observe between which two integers the sign changes.

$f(x) = x^{4}+3x^{2}-17x-115$

$f(0) = 0^{4}+3 \cdot 0^{2}-17\cdot 0-115$

$f(1) =1^{4}+3\cdot 1^{2}-17 \cdot 1-115$

$f(2) = 2^{4}+3\cdot 2^{2}-17\cdot 2-115$

$= 16+3\cdot 4 -34 -115$

$f(3) =3 ^{4}+3 \cdot 3^{2}-17 \cdot 3-115$

$= 81 + 3 \cdot 9 - 51 -115$

$f(4) = 4 ^{4}+3\cdot 4^{2}-17\cdot 4 -115$

$= 256 +3\cdot 16 -68-115$

$f(5) = 5^{4}+3 \cdot 5^{2}-17 \cdot 5 -115$

$= 625 +3 \cdot 25 -85 -115$

Since and , the -intercept is between and .

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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

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

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

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

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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