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Advanced Geometry : How to graph a function

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Example Question #51 : How To Graph A Function

Give the -coordinate(s) of the -intercept(s) of the graph of the function

$f(x) = \sqrt{x+49}$

Possible Answers:

-7, 7

Correct answer:

Explanation:

The -intercept(s) of the graph of f(x) are the point(s) at which it intersects the -axis. The -coordinate of each is 0; their -coordinate(s) are those value(s) of for which f(x) = 0 , so set up, and solve for , the equation:

f(x)= 0

The -intercept(s) of the graph of f(x) are the point(s) at which it intersects the -axis. The -coordinate of each is 0,; their -coordinate(s) are those value(s) of for which f(x) = 0 , so set up, and solve for , the equation:

f(x)= 0

$\sqrt{x+4 9} = 0$

Square both sides to eliminate the radical:

$(\sqrt{x+4 9} ) ^{2}= 0 ^{2}$

x+4 9= 0

Subtract 49 from both sides:

x+4 9 - 49 = 0-4 9

x= -49 ,

the correct choice.

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Example Question #171 : Graphing

Which of the following equations is that of an oblique asymptote of the graph of the function $f(x) = \frac{x^{3}}{x^{2}+ 2x}$ ?

Possible Answers:

y = x-1

y = x+1

y = x-2

y = x

y = x+2

Correct answer:

y = x-2

Explanation:

To find an oblique asymptote of a rational function whose numerator has higher degree than its denominator, as is the case here, divide the former by the latter, as follows:

Division

Note that "missing" terms have been inserted in the dividend as terms with zero coefficients.

Divide the leading term of the dividend by that of the divisor:

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

Place this in the quotient, and multiply this by the divisor:

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

Subtract this from the dividend. The figure should look like this:

Division

Repeat with the difference:

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

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

The figure now looks like this:

Division

The difference has degree less than that of the divisor, so the division is finished. The equation of the oblique asymptote of the graph of f(x) is taken from the quotient, and is the line of the equation y = x-2 .

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Example Question #51 : How To Graph A Function

Give the -coordinate of the -intercept of the graph of the function

$f(x)=\frac{x-3}{x^{2}+ 9}$ .

Possible Answers:

$-\frac{1}{3}$

Correct answer:

$-\frac{1}{3}$

Explanation:

The -intercept of the graph of f(x) is the point at which it intersects the -axis. Its -coordinate is 0; its -coordinate is f(0) , which can be found by substituting 0 for in the definition:

$f(x)=\frac{x-3}{x^{2}+ 9} = \frac{0-3}{0^{2}+ 9} = \frac{-3}{9} = -\frac{1}{3}$

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Example Question #51 : How To Graph A Function

Give the -coordinate(s) of the -intercept(s) of the graph of the function

$f(x) = \sqrt{x- 9}$

Possible Answers:

-9,9

The graph of f(x) has no -intercept.

-3,3

Correct answer:

Explanation:

f(x)= 0

$\sqrt{x- 9} = 0$

Square both sides to eliminate the radical:

$(\sqrt{x- 9} ) ^{2}= 0 ^{2}$

x- 9 = 0

Add 9 to both sides:

x- 9 + 9 = 0 + 9

x= 9 ,

the correct choice.

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Example Question #171 : Coordinate Geometry

Give the range of the function

$f(x) = \sqrt{x^{2}+ 25}$

Possible Answers:

The set of all real numbers

$[25, \infty)$

$[5, \infty)$

$(-\infty ,- 5 ] \cup [5, \infty)$

$(-\infty , -25 ] \cup [25, \infty)$

Correct answer:

$[5, \infty)$

Explanation:

To find the range of the function f(x) , it is necessary to note that, regardkess of the value of ,

$x ^{2} \ge 0$

Add 25 to both sides, then take the positive square root:

$x ^{2} + 25 \ge 0+ 25$

$x ^{2} + 25 \ge 25$

$\sqrt{x ^{2} + 25 }\ge \sqrt{25}$

$\sqrt{x ^{2} + 25 }\ge 5$

It follows that

$f(x) \ge 5$ ,

and that the correct range is $[5, \infty)$ .

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Example Question #171 : Coordinate Geometry

Which of the following graphs does NOT represent a function?

Possible Answers:

Act_math_159_13

Act_math_159_12

All of the graphs are functions.

Act_math_159_14

Act_math_159_10

Correct answer:

Act_math_159_13

Explanation:

This question relies on both the vertical-line test and the definition of a function. We need to use the vertical-line test to determine which of the graphs is not a function (i.e. the graph that has more than one output for a given input). The vertical-line test states that a graph represents a function when a vertical line can be drawn at every point in the graph and only intersect it at one point; thus, if a vertical line is drawn in a graph and it intersects that graph at more than one point, then the graph is not a function. The circle is the only answer choice that fails the vertical-line test, and so it is not a function.

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Advanced Geometry : How to graph a function

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #51 : How To Graph A Function

Example Question #171 : Graphing

Example Question #51 : How To Graph A Function

Example Question #51 : How To Graph A Function

Example Question #171 : Coordinate Geometry

Example Question #171 : Coordinate Geometry

All Advanced Geometry Resources

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