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Advanced Geometry : Coordinate Geometry

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Example Questions

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Example Question #11 : Graphing

Let (x,y) 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 .

Possible Answers:

$x =\frac{1}{ 1,000}$

x = 1,000

x = 100

$x =\frac{1}{ 1 00}$

The system has no solution.

Correct answer:

x = 1,000

Explanation:

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

2 u+3 v = 12

5 u- 2 v = 11

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

4 u+6 v = 24

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

19u =57

$u = 57 \div 19 = 3$

Back-solve:

2 (3)+3 v = 12

6+3 v = 12

3v = 6

v = 2

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

u = 3

$\log x = 3$

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

and

v=2

$\log y = 2$

$y=10^{2} = 100$

We check this solution in both equations:

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

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

2 (3)+3 (2)= 12

6+6 = 12 - true.

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

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

5 (3) -2 (2) = 11

15-4 = 11 - true.

(1000, 100) is the solution, and x=1,000 , the correct choice.

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Example Question #12 : Graphing

The graph of function has vertical asymptote x = 5 . Which of the following could give a definition of ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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 x-B >0 , or, equivalently, x>B and that the graph of will never cross the vertical line x=B . That makes x=B the vertical asymptote, so it follows that the graph with vertical asymptote x = 5 will have in the position. The only choice that meets this criterion is

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

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Example Question #13 : Graphing

The graph of a function has -intercept (64,0) . Which of the following could be the definition of ?

Possible Answers:

$f(x)= \log_{4}x-3$

$f(x)= \log_{2}x-6$

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

All of the other choices are correct.

$f(x)= \log_{16}x- \frac{3}{2}$

Correct answer:

All of the other choices are correct.

Explanation:

All of the functions are of the form $f(x) = \log_{b}x - C$ . To find the -intercept of such a function, we can set f(x) = 0 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 (64,0) , the equation here becomes $b^{C}= 64$ , and we can examine each of the functions by finding the value of $b^{C}$ .

$f(x)= \log_{8}x-2$ :

b= 8, c= 2

$b^{C}= 8^{2} = 64$

$f(x)= \log_{2}x-6$ :

b = 2,c=6

$b^{C}= 2^{6} = 64$

$f(x)= \log_{4}x-3$ :

b=4,c=3

$b^{C}=4^{3} = 64$

$f(x)= \log_{16}x- \frac{3}{2}$ :

$b=16,c=\frac{3}{2}$

$b^{C}= 16^{ \frac{3}{2}} = \left ( \sqrt{16} \right )^{3}= 4^{3}= 64$

All four choices fit the criterion.

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Example Question #14 : Graphing

The graph of a function has $y\,$ -intercept (0,8) . Which of the following could be the definition of ?

Possible Answers:

$f(x)= \log _{2}(x+1) +6$

$f(x)= \log _{2}(x+8) +6$

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

$f(x)= \log _{2}(x+4) +6$

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

Correct answer:

$f(x)= \log _{2}(x+4) +6$

Explanation:

All of the functions take the form

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

for some integer . To find the choice that has $y\,$ -intercept (0,8) , set x = 0 and f(x)= 8 , 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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Example Question #15 : Graphing

Define a function as follows:

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

Give the -intercept of the graph of .

Possible Answers:

$\left ( -2 - 16 \sqrt{2}, 0 \right )$

(2,0)

(6,0)

$\left ( -2 + 16 \sqrt{2}, 0 \right )$

(-2, 0)

Correct answer:

(2,0)

Explanation:

Set g(x) =0 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:

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

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

x=2 .

The -intercept is (2,0) .

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Example Question #16 : Graphing

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.

Possible Answers:

$\log 1\frac{5}{9}$

$2 \frac{4}{9}$

$\log 2\frac{34}{81}$

The graphs of and do not intersect.

Correct answer:

The graphs of and do not intersect.

Explanation:

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

f(x)= g(x)

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

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

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

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

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

-9x = - 22

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

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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Example Question #11 : How To Graph A Logarithm

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

Possible Answers:

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

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

None of the other responses gives a correct answer.

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

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

Correct answer:

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

Explanation:

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 f(x) = 0 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 = 5, C=2

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

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

b = 2, C=5

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

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

b = 5, C=-2

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

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

b = 2, C=-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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Example Question #11 : Coordinate Geometry

Define a function as follows:

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

A line passes through the - and $y\,$ -intercepts of the graph of . Give the equation of the line.

Possible Answers:

y =- 12x+30

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

y = 12x+24

$y= -\frac{1}{12} x -4$

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

Correct answer:

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

Explanation:

The -intercept of the graph of can befound by setting g(x) = 0 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:

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

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

$x+3 = 3^{3}$

x+3 = 27

x = 24

The -intercept of the graph of is (24,0) .

The $y \,$ -intercept of the graph of can be found by evaluating g(0):

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

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

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

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

$= \log_{9}9- 3$

= 1-3

= -2

The $y \,$ -intercept of the graph of is (0, -2) .

If (a,0) and (0,b) are the - and $y\,$ -intercepts, respectively, of a line, the slope of the line is $m=-\frac{b}{a}$ . Substituting a = 24 and b = -2 , this is

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

Setting $m= \frac{1}{12}$ and b = -2 in the slope-intercept form of the equation of a line:

y = mx+b

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

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Example Question #19 : Graphing

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.

Possible Answers:

$\log 6$

The graphs of and do not intersect.

$\log 9$

Correct answer:

$\log 9$

Explanation:

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

f(x)= g(x)

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

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:

(x+2)(x+6 ) = 0

Either x+2 = 0 , in which case x=-2

x+6 = 0 , in which case x = -6

Note, however, that we can eliminate x = -6 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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Example Question #11 : Graphing

This graph shows the graph of $y=\log _{2}x$ . The blue point has an -coordinate of . What is the -coordinate?

Log base 2

Possible Answers:

256

Correct answer:

Explanation:

This point has an x-coordinate of 8, so we can figure out what y is just by plugging in 8 for x:

$y=\log_{2}(8)$

This can be evaluated using a calculator, or just by understanding what a logarithm means.

This is essentially asking "2 to what power gives us 8," which is 3.

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Advanced Geometry : Coordinate Geometry

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #11 : Graphing

Example Question #12 : Graphing

Example Question #13 : Graphing

Example Question #14 : Graphing

Example Question #15 : Graphing

Example Question #16 : Graphing

Example Question #11 : How To Graph A Logarithm

Example Question #11 : Coordinate Geometry

Example Question #19 : Graphing

Example Question #11 : Graphing

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