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Advanced Geometry : Advanced Geometry

Study concepts, example questions & explanations for Advanced Geometry

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Example Questions

Example Question #1 : Coordinate Geometry

What is the -intercept of the graph of $y = \log _{2}(x+ 4)$ ?

Possible Answers:

The graph has no -intercept.

(-3,0)

(1,0)

(2,0)

(-4,0)

Correct answer:

(-3,0)

Explanation:

Set y = 0 and solve:

$y = \log _{2}(x+ 4)$

$\log _{2}(x+ 4) = 0$

$2 ^ {\log _{2}(x+ 4) }= 2 ^ { 0}$

x + 4 = 1

x + 4 - 4 = 1 - 4

x = -3

The -intercept is (-3,0) .

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Example Question #1 : Coordinate Geometry

What is the -intercept of the graph of $y = \log _{3}(x+ 9)$ ?

Possible Answers:

(0,2)

The graph has no -intercept.

(0,3)

(0,-6)

(0,-9)

Correct answer:

(0,2)

Explanation:

Set x= 0 and evaluate :

$y = \log _{3}(x+ 9)$

$y = \log _{3}(0+ 9) = \log _{3}9$

Since $3 ^{2} =9$ ,

$\log _{3}9 =2$ , and the -intercept is (0,2) .

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Example Question #1 : Coordinate Geometry

What is the vertical asymptote of the graph of $y = \log_{5} (x-7)$ ?

Possible Answers:

The graph has no vertical asymptote.

x = 12

x = 6

x = 7

x = 8

Correct answer:

x = 7

Explanation:

The graph of a logarithmic function has a vertical asymptote which can be found by finding the value at which the power is equal to 0:

$x-7 = 0 \Rightarrow x = 7$

If $x \leq 7$ , then $\log_{5} (x-7)$ is an undefined expression, so the vertical asymptote is x = 7 .

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Example Question #1 : Graphing A Logarithm

Define a function as follows:

$g(x) =2 \log_{4}(x-4)+ 3$

Give the -intercept of the graph of .

Possible Answers:

The graph of has no -intercept.

$\left ( 3\frac{7}{8}, 0 \right )$

$\left ( 4\frac{1}{8}, 0 \right )$

(12,0)

(-4, 0)

Correct answer:

$\left ( 4\frac{1}{8}, 0 \right )$

Explanation:

Set g(x) =0 and evaluate to find the -coordinate of the -intercept.

$g(x) =2 \log_{4}(x-4)+ 3 = 0$

$2 \log_{4}(a-4) = -3$

$\log_{4}(a-4) = -\frac{3}{2}$

This can be rewritten in exponential form:

$a - 4 = 4 ^{-\frac{3}{2}}= \frac{1}{4 ^{ \frac{3}{2}}} = \frac{1}{(\sqrt{4 })^{ 3}} = \frac{1}{2^{ 3}} = \frac{1}{8}$

$a = 4\frac{1}{8}$

The -intercept of the graph of is $\left ( 4\frac{1}{8}, 0 \right )$ .

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Example Question #1 : Coordinate Geometry

Define a function as follows:

$g(x) = \log_{9}(x-4)+ 2$

Give the $y\:$ -intercept of the graph of .

Possible Answers:

The graph of has no $y\:$ -intercept.

(0,4)

$\left ( 0, 4\frac{1}{81}\right )$

(0, 2)

$\left ( 0, 3\frac{80}{81}\right )$

Correct answer:

The graph of has no $y\:$ -intercept.

Explanation:

The $y\:$ -coordinate of the $y\:$ -intercept is g(0) :

$g(x) = \log_{9}(x-4)+ 2$

$g(0) = \log_{9}(0-4)+ 2$

$= \log_{9}( -4)+ 2$

However, the logarithm of a negative number is an undefined expression, so g(0) is an undefined quantity, and the graph of has no $y\:$ -intercept.

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Example Question #5 : Coordinate Geometry

Define a function as follows:

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

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

Possible Answers:

x = -3

x = 3

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

x = -2

x = 2

Correct answer:

x = -2

Explanation:

Only positive numbers have logarithms, so

x+2 > 0

x> -2

The graph never crosses the vertical line of the equation x = -2 , so this is the vertical asymptote.

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Example Question #6 : Coordinate Geometry

Define a function as follows:

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

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

Possible Answers:

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

x = 2

x = 4

x = 3

The graph of has no vertical asymptote.

Correct answer:

x = 4

Explanation:

Only positive numbers have logarithms, so

x- 4 > 0

x> 4

The graph never crosses the vertical line of the equation x = 4 , so this is the vertical asymptote.

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Example Question #7 : Coordinate Geometry

Define a function as follows:

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

Give the $y\:$ -intercept of the graph of .

Possible Answers:

The graph of has no $y\:$ -intercept.

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

$\left ( 2\frac{1}{3},0 \right )$

(9,0)

(-3,0)

Correct answer:

$\left ( 2\frac{1}{3},0 \right )$

Explanation:

The $y\:$ -coordinate of the $y\:$ -intercept is g(0) :

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

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

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

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

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Example Question #1 : Coordinate Geometry

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.

Possible Answers:

$\log 2$

$-\log 2$

The graphs of and do not intersect.

Correct answer:

$\log 2$

Explanation:

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

g(x)= f(x)

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

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:

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

2x (2x+5)+7(2x+5 )=0

(2x +7)(2x+5 )=0

2x+7 = 0

2x= -7

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

2x+5= 0

2x= -5

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

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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Example Question #1 : Graphing A Logarithm

Let (x,y) 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 $y\:$ .

Possible Answers:

$y = \frac{1}{27}$

y = 9

$y = \frac{1}{9}$

$y = \frac{1}{27}$

y = -27

Correct answer:

$y = \frac{1}{27}$

Explanation:

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

2 u-3 v = 13

5 u+2 v = 4

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

4 u-6 v = 26

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

19u =38

$u = 38 \div 19 = 2$

Now back-solve:

5 (2)+2 v = 4

10+2 v = 4

2v=-6

v= -3

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

u = 2

$\log_{2} x = 2$

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

v= -3

$\log_{3} y = -3$

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

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

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Advanced Geometry : Advanced Geometry

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #1 : Coordinate Geometry

Example Question #1 : Coordinate Geometry

Example Question #1 : Coordinate Geometry

Example Question #1 : Graphing A Logarithm

Example Question #1 : Coordinate Geometry

Example Question #5 : Coordinate Geometry

Example Question #6 : Coordinate Geometry

Example Question #7 : Coordinate Geometry

Example Question #1 : Coordinate Geometry

Example Question #1 : Graphing A Logarithm

All Advanced Geometry Resources

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