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Advanced Geometry : How to graph a function

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Example Questions

Example Question #1 : How To Graph A Function

The chord of a $60^{\circ }$ central angle of a circle with area $40 \pi$ has what length?

Possible Answers:

$10\sqrt{2}$

$4 \sqrt{5}$

$2 \sqrt{10}$

Correct answer:

$2 \sqrt{10}$

Explanation:

The radius of a circle with area $40 \pi$ can be found as follows:

$\pi r^{2} = A$

$\pi r^{2} = 40 \pi$

$r^{2} = 40$

$r = \sqrt{40 } = \sqrt{4} \cdot \sqrt{10} = 2 \sqrt{10}$

The circle, the central angle, and the chord are shown below:

Chord

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved equilateral, so $AB = AO= 2 \sqrt{10}$ , the correct response.

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Example Question #2 : How To Graph A Function

The chord of a $120^{\circ }$ central angle of a circle with area $32 \pi$ has what length?

Possible Answers:

$4\sqrt{3}$

$4 \sqrt{2}$

$4 \sqrt{6}$

Correct answer:

$4 \sqrt{6}$

Explanation:

The radius of a circle with area $32 \pi$ can be found as follows:

$\pi r^{2} = A$

$\pi r^{2} = 32 \pi$

$r^{2} = 32$

$r = \sqrt{32} = \sqrt{16} \cdot \sqrt{2} = 4 \sqrt{2}$

The circle, the central angle, and the chord are shown below, along with $\overline{AM}$ , which bisects isosceles $\Delta AOB$

Chord

We concentrate on $\Delta AOM$ , a 30-60-90 triangle. By the 30-60-90 Theorem,

$OM = \frac{1}{2}AO = \frac{1}{2} \cdot 4 \sqrt{2} = 2\sqrt{2}$

and

$AM = OM \sqrt{3} = 2 \sqrt{2} \cdot \sqrt{3} = 2 \sqrt{6}$

The chord $\overline{AB}$ has length twice this, or

$2 \cdot 2 \sqrt{6} = 4 \sqrt{6}$

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Example Question #121 : Coordinate Geometry

The chord of a $60^{\circ }$ central angle of a circle with circumference $90 \pi$ has what length?

Possible Answers:

$45\sqrt{3}$

$90\sqrt{2}$

$45\sqrt{2}$

Correct answer:

Explanation:

A circle with circumference $90 \pi$ has as its radius

$90\pi \div 2\pi = 45$ .

The circle, the central angle, and the chord are shown below:

Chord

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved equilateral, so AB = AO= 45 , the correct response.

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Example Question #1041 : Problem Solving Questions

What is the domain of $y = 4 - x^{2}$ ?

Possible Answers:

$x \geq 4$

all real numbers

$x \leq 4$

$x \leq 0$

x=0

Correct answer:

all real numbers

Explanation:

The domain of the function specifies the values that can take. Here, $4-x^{2}$ is defined for every value of , so the domain is all real numbers.

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Example Question #1041 : Gmat Quantitative Reasoning

What is the domain of $y=-2\sqrt{x}$ ?

Possible Answers:

$x\geq 0$

x>0

$x\leq 0$

x<-2

x=0

Correct answer:

$x\geq 0$

Explanation:

To find the domain, we need to decide which values can take. The is under a square root sign, so cannot be negative. can, however, be 0, because we can take the square root of zero. Therefore the domain is $x\geq 0$ .

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Example Question #62 : Graphing

What is the domain of the function $y=\sqrt{4-x^{2}}$ ?

Possible Answers:

$x\leq -2$

$\left [ -2,2 \right ]$

x>-2

x<-2

$\left ( -2,2 \right )$

Correct answer:

$\left [ -2,2 \right ]$

Explanation:

To find the domain, we must find the interval on which $\sqrt{4-x^{2}}$ is defined. We know that the expression under the radical must be positive or 0, so $\sqrt{4-x^{2}}$ is defined when $x^{2}\leq 4$ . This occurs when $x \geq -2$ and $x \leq 2$ . In interval notation, the domain is $\left [ -2,2 \right ]$ .

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Example Question #1042 : Gmat Quantitative Reasoning

Define the functions and as follows:

$\small \small f (x) = 2x + \sqrt{x - 1}$

$\small g (x) = 6x - \sqrt{x-1}$

What is the domain of the function (f+g)(x) ?

Possible Answers:

$\small \small (-\infty , -1] \cup [1, \infty )$

$\small (-\infty , 1) \cup (1, \infty )$

$\small (-\infty ,\infty )$

$\small (1, \infty )$

$\small \small [1, \infty )$

Correct answer:

$\small \small [1, \infty )$

Explanation:

The domain of (f+g) is the intersection of the domains of and . and are each restricted to all values of that allow the radicand x-1 to be nonnegative - that is,

$\small \small x - 1\geq 0$ , or

$\small \small \small x \geq 1$

Since the domains of and are the same, the domain of (f+g) is also the same. In interval form the domain of (f+g) is $\small \small [1, \infty )$

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Example Question #4 : How To Graph A Function

Define $g (x) = \frac{1}{\sqrt[3]{x -27}}$

What is the natural domain of ?

Possible Answers:

$(-\infty , - 27)\cup (27, \infty )$

$(-\infty , - 3)\cup (3, \infty )$

$(-\infty , 27)\cup (27, \infty )$

$(3, \infty )$

$(27, \infty )$

Correct answer:

$(-\infty , 27)\cup (27, \infty )$

Explanation:

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression $\sqrt[3]{x -27}$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which

$\sqrt[3]{x -27} = 0$

$\left (\sqrt[3]{x -27} \right )^{3}= 0^{3}$

x -27 = 0

x = 27

27 is the only number excluded from the domain.

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Example Question #5 : Graphing A Function

Define $g(x) = \frac{x}{x^{2}+3x-4 }$

What is the natural domain of ?

Possible Answers:

$( - \infty ,-1 ) \cup (-1,4) \cup (4, \infty )$

$( - \infty ,-1 ) \cup (-1,0) \cup (0,4) \cup (4, \infty )$

$( - \infty ,-4 ) \cup (-4,0) \cup (0,1) \cup (1, \infty )$

$( - \infty ,-4 ) \cup (-4,1) \cup (1, \infty )$

$( - \infty ,0) \cup (0, \infty )$

Correct answer:

$( - \infty ,-4 ) \cup (-4,1) \cup (1, \infty )$

Explanation:

Since the expression $x^{2}+3x-4$ is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which $x^{2}+3x-4 =0$ . We solve for by factoring the polynomial, which we can do as follows:

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

(x+?)(x+?)=0

Replacing the question marks with integers whose product is and whose sum is 3:

(x+4)(x-1)=0

$x + 4 = 0 \Rightarrow x = -4$

$x -1 = 0 \Rightarrow x = 1$

Therefore, the domain excludes these two values of .

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Example Question #1 : How To Graph A Function

Define $f (x) = \frac{1}{x ^{2}- 25 }$ .

What is the natural domain of ?

Possible Answers:

$(-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)$

$(-\infty , -5 ) \cup \left ( -5, \infty)$

(-5 , 5)

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

$(-\infty , 5 ) \cup \left ( 5, \infty)$

Correct answer:

$(-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)$

Explanation:

The only restriction on the domain of is that the denominator cannot be 0. We set the denominator to 0 and solve for to find the excluded values:

$x ^{2}-25 =0$

$x ^{2}= 25$

$x = -5 \textrm{ or }x = 5$

The domain is the set of all real numbers except those two - that is,

$(-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)$ .

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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 #1 : How To Graph A Function

Example Question #2 : How To Graph A Function

Example Question #121 : Coordinate Geometry

Example Question #1041 : Problem Solving Questions

Example Question #1041 : Gmat Quantitative Reasoning

Example Question #62 : Graphing

Example Question #1042 : Gmat Quantitative Reasoning

Example Question #4 : How To Graph A Function

Example Question #5 : Graphing A Function

Example Question #1 : How To Graph A Function

All Advanced Geometry Resources

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