How to graph a function - Advanced Geometry

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Question

Which of the following graphs does NOT represent a function?

Answer

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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Question

Suppose .

To obtain the graph of , shift the graph a distance of units .

Answer

There are four shifts of the graph y = f(x):

y = f(x) + c shifts the graph c units upwards.

y = f(x) – c shifts the graph c units downwards.

y = f(x + c) shifts the graph c units to the left.

y = f(x – c) shifts the graph c units to the right.

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Question

The figure above shows the graph of y = f(x). Which of the following is the graph of y = |f(x)|?

Answer

One of the properties of taking an absolute value of a function is that the values are all made positive. The values themselves do not change; only their signs do. In this graph, none of the y-values are negative, so none of them would change. Thus the two graphs should be identical.

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Question

Below is the graph of the function :

Which of the following could be the equation for ?

Answer

First, because the graph consists of pieces that are straight lines, the function must include an absolute value, whose functions usually have a distinctive "V" shape. Thus, we can eliminate f(x) = x2 – 4x + 3 from our choices. Furthermore, functions with x2 terms are curved parabolas, and do not have straight line segments. This means that f(x) = |x2 – 4x| – 3 is not the correct choice.

Next, let's examine f(x) = |2x – 6|. Because this function consists of an abolute value by itself, its graph will not have any negative values. An absolute value by itself will only yield non-negative numbers. Therefore, because the graph dips below the x-axis (which means f(x) has negative values), f(x) = |2x – 6| cannot be the correct answer.

Next, we can analyze f(x) = |x – 1| – 2. Let's allow x to equal 1 and see what value we would obtain from f(1).

f(1) = | 1 – 1 | – 2 = 0 – 2 = –2

However, the graph above shows that f(1) = –4. As a result, f(x) = |x – 1| – 2 cannot be the correct equation for the function.

By process of elimination, the answer must be f(x) = |2x – 2| – 4. We can verify this by plugging in several values of x into this equation. For example f(1) = |2 – 2| – 4 = –4, which corresponds to the point (1, –4) on the graph above. Likewise, if we plug 3 or –1 into the equation f(x) = |2x – 2| – 4, we obtain zero, meaning that the graph should cross the x-axis at 3 and –1. According to the graph above, this is exactly what happens.

The answer is f(x) = |2x – 2| – 4.

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Question

Which of the following could be a value of $f(x)$ for $f(x)=-x^2 + 3$ ?

Answer

The graph is a down-opening parabola with a maximum of $y=3$ . Therefore, there are no y values greater than this for this function.

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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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 ?

Answer

The domain of is the intersection of the domains of and . and are each restricted to all values of that allow the radicand 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 is also the same. In interval form the domain of is $\small \small [1, \infty )$

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Question

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

What is the natural domain of ?

Answer

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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Question

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

What is the natural domain of ?

Answer

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

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

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

27 is the only number excluded from the domain.

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Question

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

What is the natural domain of ?

Answer

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:

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

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

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

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

Therefore, the domain excludes these two values of .

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Question

What is the equation for the line pictured above?

Answer

A line has the equation

where is the intercept and is the slope.

The intercept can be found by noting the point where the line and the y-axis cross, in this case, at so .

The slope can be found by selecting two points, for example, the y-intercept and the next point over that crosses an even point, for example, .

Now applying the slope formula,

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

which yields $m=\frac{-1-2}{2-0}=-\frac{3}{2}$ .

Therefore the equation of the line becomes:

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

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Question

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

Answer

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:

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

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Question

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

Answer

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:

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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Question

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

Answer

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 $\Delta AOB$

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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Question

Which of the following graphs represents the y-intercept of this function?

Answer

Graphically, the y-intercept is the point at which the graph touches the y-axis. Algebraically, it is the value of when .

Here, we are given the function . In order to calculate the y-intercept, set equal to zero and solve for .

So the y-intercept is at .

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Question

Which of the following graphs represents the x-intercept of this function?

Answer

Graphically, the x-intercept is the point at which the graph touches the x-axis. Algebraically, it is the value of for which .

Here, we are given the function . In order to calculate the x-intercept, set equal to zero and solve for .

So the x-intercept is at .

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Question

Which of the following represents $f(x)=\frac{1}{2}x-2$ ?

Answer

A line is defined by any two points on the line. It is frequently simplest to calculate two points by substituting zero for x and solving for y, and by substituting zero for y and solving for x.

$f(x)=\frac{1}{2}x-2$

Let . Then

$y=\frac{1}{2}(0)-2$

So our first set of points (which is also the y-intercept) is

Let . Then

$0=\frac{1}{2}x-2$

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

So our second set of points (which is also the x-intercept) is .

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Question

Which graph accurately represents the following function:

Answer

The first step in determining which graph is correct is finding the origin of the function. If both x and y are equal to 0, the coordinates of the origin would be . The second step is to determine whether the graph opens up or down. The x and y are both positive, so the parabola will open upwards. The correct graph will look like

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