Coordinate Geometry - GMAT Quantitative

Card 0 of 2080

Question

Define a function as follows:

$g(x) =2 \ln (x-4)+ 3$

Give the equation of the vertical asymptote of the graph of .

Answer

Only positive numbers have logarithms, so

The graph never crosses the vertical line of the equation , so this is the vertical asymptote.

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Question

Define a function as follows:

$g(x) =-2 \log_{8}(x+2)+ 3$

Give the -intercept of the graph of .

Answer

The -coordinate of the -intercept is :

$g(x) =-2 \log_{8}(x+2)+ 3$

$g(0) =-2 \log_{8}(0 +2)+ 3$

$g(0) =-2 \log_{8}2+ 3$

Since 2 is the cube root of 8, $8^{\frac{1}{3}}= 2$ , and $\log_{8}2 = \frac{1}{3}$ . Therefore,

$g(0) =-2 \cdot \frac{1}{3}+ 3 = -\frac{2}{3}+ 3 = 2\frac{1}{3}$ .

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

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Question

Define a function as follows:

$g(x) =-3 \log_{8}(x+2)+ 2$

Give the -intercept of the graph of .

Answer

Set and evaluate to find the -coordinate of the -intercept.

$g(x) =-3 \log_{8}(x+2)+ 2 = 0$

$-3 \log_{8}(x+2)= - 2$

$\log_{8}(x+2)= \frac{-2}{-3}= \frac{2}{3}$

Rewrite in exponential form:

$x+ 2 = 8^{\frac{2}{3}}$

$x+ 2 =\left ( \sqrt[3]{8} \right )^{2} = 2^{2} = 4$

.

The -intercept is .

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Question

Define functions and as follows:

$f(x) = 2 \log (x+5)$

$g(x) = \log (2x+13)$

Give the -coordinate of a point at which the graphs of the functions intersect.

Answer

Since $a \log N= \log( N^{a})$ , the definition of can be rewritten as follows:

$f(x) = \log \left [(x+5)^{2} \right ]$ .

Find the -coordinate of the point at which the graphs of and meet by setting

$\log \left [(x+5)^{2} \right ]= \log (2x+13)$

Since the common logarithms of the two polynomials are equal, we can set the polynomials themselves equal, then solve:

$(x+5)^{2} = 2x+13$

$x^{2} +10x+25= 2x+13$

$x^{2} +10x+25- 2x - 13 = 2x+13 - 2x - 13$

$x^{2} +8x+12= 0$

The quadradic trinomial can be "reverse-FOILed" by noting that 2 and 6 have product 12 and sum 8:

Either , in which case

or

, in which case

Note, however, that we can eliminate as a possible -value, since

$f(-6) = 2 \log (-6+5) = 2 \log (-1)$ ,

an undefined quantity since negative numbers do not have logarithms.

Since

$f(-2) = 2 \log (-2+5) = 2 \log 3 = \log (3^{2}) = \log 9$

and

$g(-2) = \log [2(-2)+13] = \log (-4+13) = \log 9$ ,

is the correct -value, and $\log 9$ is the correct -value.

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Question

Define functions and as follows:

$f(x) = 2 \log (x-4)$

$g(x) = \log (x+3)+ \log (x-2)$

Give the -coordinate of a point at which the graphs of the functions intersect.

Answer

Since $a \log N = \log( N^{a})$ , the definition of can be rewritten as follows:

$f(x) = 2 \log (x-4)$

$f(x) = \log (x-4) ^{2}$

Since $\log M + \log N = \log MN$ , the definition of can be rewritten as follows:

$g(x) = \log (x+3)+ \log (x-2)$

$g(x) = \log (x+3) (x-2)$

First, we need to find the -coordinate of the point at which the graphs of and meet by setting

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

Since the common logarithms of the two polynomials are equal, we can set the polynomials themselves equal, then solve:

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

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

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

$x = \frac{22}{9} = 2\frac{4}{9}$

However, if we evaluate $f \left ( 2\frac{4}{9} \right )$ , the expression becomes

$f(x) = 2 \log (x-4)$

$f\left ( 2\frac{4}{9} \right ) = 2\log \left ( 2\frac{4}{9} -4\right ) = 2 \log \left ( -1\frac{5}{9} \right )$ ,

which is undefined, since a negative number cannot have a logarithm.

Consequently, the two graphs do not intersect.

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Question

Define functions and as follows:

$f(x) = \log \frac{1}{x+3}$

$g(x) = \log (4x+12)$

Give the -coordinate of a point at which the graphs of the functions intersect.

Answer

Since $- \log N = \log \left ( \frac{1}{N} \right )$ , the definition of can be rewritten as follows:

$f(x) = - \log (x+3)$

First, we need to find the -coordinate of the point at which the graphs of and meet by setting

$\log (4x+12)= \log \frac{1}{x+3}$

Since the common logarithms of the polynomial and the rational expression are equal, we can set those expressions themselves equal, then solve:

$(4x+12)= \frac{1}{x+3}$

$(4x+12)(x+3)= \frac{1}{x+3} \cdot (x+3)$

$4x^{3}+24x+36 =1$

$4x^{3}+24x+35 =0$

We can solve using the method, finding two integers whose sum is 24 and whose product is $4 \cdot 35 = 140$ - these integers are 10 and 14, so we split the niddle term, group, and factor:

$(4x^{3}+10x)+(14x+35 )=0$

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

or

$x = -\frac{5}{2} = -2\frac{1}{2}$

This gives us two possible -coordinates. However, since

$g\left ( -3\frac{1}{2} \right ) = \log \left [4\left ( -3\frac{1}{2} \right ) +12 \right ] = \log (-14+12) = \log (-2)$ ,

an undefined quantity - negative numbers not having logarithms -

we throw this value out. As for the other -value, we evaluate:

$f\left ( -2\frac{1}{2} \right ) = \log \frac{1}{ -2\frac{1}{2} +3} = \log \frac{1}{\frac{1}{2}} = \log 2$

and

$g\left ( -2\frac{1}{2} \right ) = \log \left [4\left ( -2\frac{1}{2} \right ) +12 \right ] = \log (-10+12) = \log 2$

$-2 \frac{1}{2}$ is the correct -value, and $\log 2$ is the correct -value.

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Question

Define a function as follows:

$g(x) =2 \log_{9}(x+3)- 3$

A line passes through the - and -intercepts of the graph of . Give the equation of the line.

Answer

The -intercept of the graph of can befound by setting and solving for :

$2 \log_{9}(x+3)- 3 = 0$

$2 \log_{9}(x+3) = 3$

$\log_{9}(x+3) = \frac{3}{2}$

Rewritten in exponential form:

$x+3 = 9 ^{\frac{3}{2}}$

$x+3 = (\sqrt{9} )^{3}$

$x+3 = 3^{3}$

The -intercept of the graph of is .

The -intercept of the graph of can be found by evaluating

$g(x) =2 \log_{9}(x+3)- 3$

$g(0) =2 \log_{9}(0+3)- 3$

$=2 \log_{9}3- 3$

$= \log_{9}3^{2}- 3$

$= \log_{9}9- 3$

The -intercept of the graph of is .

If and are the - and -intercepts, respectively, of a line, the slope of the line is $m=-\frac{b}{a}$ . Substituting and , this is

$m=-\frac{-2}{24} = \frac{1}{12}$ .

Setting $m= \frac{1}{12}$ and in the slope-intercept form of the equation of a line:

$y= \frac{1}{12} x -2$

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Question

Let be the point of intersection of the graphs of these two equations:

$2 \log_{2} x-3 \log_{3} y = 13$

$5 \log_{2} x+2 \log_{3} y = 4$

Evaluate .

Answer

Substitute and for $\log_{2} x$ and $\log_{3} y$ , respectively, and solve the resulting system of linear equations:

Multiply the first equation by 2, and the second by 3, on both sides, then add:

$\underline{15 u+6 v = 12}$

$u = 38 \div 19 = 2$

Now back-solve:

We need to find both and to ensure a solution exists. By substituting back:

$\log_{2} x = 2$

$x = 2^{2} = 4$ .

$\log_{3} y = -3$

$y = 3^{-3} = \frac{1}{3^{3}} = \frac{1}{27}$

$\left (4, \frac{1}{27} \right )$ is the solution, and $y = \frac{1}{27}$ , the correct choice.

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Question

Let be the point of intersection of the graphs of these two equations:

$2 \log x+3 \log y = 12$

$5 \log x- 2 \log y = 11$

Evaluate .

Answer

Substitute and for $\log x$ and $\log y$ , respectively, and solve the resulting system of linear equations:

Multiply the first equation by 2, and the second by 3, on both sides, then add:

$\underline{15 u- 6 v = 33}$

$u = 57 \div 19 = 3$

Back-solve:

We need to find both and to ensure a solution exists. By substituting back:

$\log x = 3$

$x=10^{3} = 1,000$

and

$\log y = 2$

$y=10^{2} = 100$

We check this solution in both equations:

$2 \log x+3 \log y = 12$

$2 \log 1,000+3 \log 100 = 12$

- true.

$5 \log x- 2 \log y = 11$

$5 \log 1,000- 2 \log 100 = 11$

- true.

is the solution, and , the correct choice.

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Question

The graph of function has vertical asymptote . Which of the following could give a definition of ?

Answer

Given the function $f(x) = A \log (x-B)+C$ , the vertical asymptote can be found by observing that a logarithm cannot be taken of a number that is not positive. Therefore, it must hold that , or, equivalently, and that the graph of will never cross the vertical line . That makes the vertical asymptote, so it follows that the graph with vertical asymptote will have in the position. The only choice that meets this criterion is

$f(x) = 3 \log (x-5)+7$

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Question

The graph of a function has -intercept . Which of the following could be the definition of ?

Answer

All of the functions take the form

$f(x)= \log _{2}(x+A) +6$

for some integer . To find the choice that has -intercept , set and , and solve for :

$8= \log _{2}(0+A) +6$

$8= \log _{2}A +6$

$2= \log _{2}A$

In exponential form:

$A =2^{2}=4$

The correct choice is $f(x)= \log _{2}(x+4) +6$ .

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Question

The graph of a function has -intercept $\left ( \frac{1}{32}, 0 \right )$ . Which of the following could be the definition of ?

Answer

All of the functions are of the form $f(x) = \log_{b}x - C$ . To find the -intercept of a function $f(x) = \log_{b}x - C$ , we can set and solve for :

$\log_{b}x - C = 0$

$\log_{b}x =C$

$b^{C}= x$ .

Since we are looking for a function whose graph has -intercept $\left ( \frac{1}{32}, 0 \right )$ , the equation here becomes $b^{C}= \frac{1}{32}$ , and we can examine each of the functions by finding the value of $b^{C}$ and seeing which case yields this result.

$f(x)= \log_{5}x-2$ :

$b^{C} = 5^{2} = 25$

$f(x)= \log_{2}x-5$ :

$b^{C} = 2^{5} = 32$

$f(x)= \log_{5}x+2$ :

$b^{C} = 5^{-2} = \frac{1}{5^{2}}= \frac{1}{25}$

$f(x)= \log_{2}x+5$ :

$b^{C} = 2^{-5} = \frac{1}{2^{5}}= \frac{1}{32}$

The graph of $f(x)= \log_{2}x+5$ has -intercept $\left ( \frac{1}{32}, 0 \right )$ and is the correct choice.

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Question

The point lies on a line with slope that passes through the point .

What is ?

Answer

We first need to find an equation of the line with a slope of that passes through the point (2, 5).

Now, plug in the point and solve for .

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Question

A line goes through points . What is ?

Answer

The slope of the line through and can be found using the slope formula:

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

Set $x_{1}=1,x_{2} = 4, y_{1}=4,y_{2} = 8$ :

$m = \frac{8-4}{4-1} = \frac{4}{3}$

We can use this slope and the slope formula; set $x_{1}=1,x_{2} = -3, y_{1}=4,y_{2} = N$ :

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

$\frac{N-4}{-3-1} = \frac{4}{3}$

$\frac{N-4}{-4} = \frac{4}{3}$

$\frac{N-4}{-4}\cdot \left ( -4\right ) = \frac{4}{3} \cdot \left ( -4\right )$

$N-4= -\frac{16}{3}$

$N-4+4= -\frac{16}{3} +4$

$N= -\frac{4}{3} = -1 \frac{1}{3}$

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Question

The point lies on a line with a slope that passes through . What is the value of ?

Answer

In order to find the value of , we first need to find the equation for the line with slope that passes through the point .

Plugging in and solving for :

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Question

The point lies on a line with a slope $m=\frac{1}{2}$ that passes through . What is the value of ?

Answer

In order to find the value of , we first need to find the equation for the line with a slope $m=\frac{1}{2}$ that passes through .

$y-6=\frac{1}{2}(x-3)$

$y-6=\frac{1}{2}x-\frac{3}{2}$

$y=\frac{1}{2}x+\frac{9}{2}$

Plugging in and solving for :

$d=\frac{1}{2}(2)+\frac{9}{2}$

$d=\frac{11}{2}$

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Question

The point lies on a line with a slope that passes through . What is the value of ?

Answer

In order to find the value of , we first need to find the equation for the line with a slope that passes through .

Plugging in and solving for :

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Question

What are the possible values of if the parabola of the quadratic function $f (x) = Ax^{2} + 4x- 2$ is concave upward and does not intersect the -axis?

Answer

If the graph of $f (x) = Ax^{2} + 4x- 2$ is concave upward, then .

If the graph does not intersect the -axis, then $Ax^{2} + 4x- 2 = 0$ has no real solution, and the discriminant $4^{2} - 4 \cdot A \cdot \left ( -2\right )$ is negative:

$4^{2} - 4 \cdot A \cdot \left ( -2\right ) < 0$

$8 A \div 8< - 16 \div 8$

For the parabola to have both characteristics, it must be true that and , but these two events are mutually exclusive. Therefore, the parabola cannot exist.

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Question

Which of the following is the equation of the line of symmetry of a horizontal parabola on the coordinate plane with its vertex at ?

Answer

The line of symmetry of a horizontal parabola is a horizontal line, the equation of which takes the form for some . The line of symmetry passes through the vertex, which here is , so the equation must be .

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Question

Which of the following is the equation of the line of symmetry of a vertical parabola on the coordinate plane with its vertex at ?

Answer

The line of symmetry of a vertical parabola is a vertical line, the equation of which takes the form for some . The line of symmetry passes through the vertex, which here is , so the equation must be .

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