Card 0 of 928
In the standard xy coordinate plane, what quadrant(s) could point A lie in, if its x-coordinate and y-coordinate have opposite signs?
We know that right of the origin represents the positive x-direction, and above the origin represents the positive y-direction. For Quadrant I, both the x-coordinate and y-coordinate must be positive, i.e. (+,+). For quadrant II, the x-coordinate is negative, while the y-coordinate is positive, i.e. (+, -). For Quadrant III, both coordinates must be negative, i.e. (-, -). For Quadrant IV, the x-coordinate is positive, and the y-coordinate is negative, i.e. (+, -).
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The point on the coordinate plane with coordinates lies _____
On the coordinate plane, a point with a negative -coordinate and a negative
-coordinate lies in the lower left quadrant - Quadrant III.
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Which of the following ordered pairs is in Quadrant III?
By definition, only Quadrant III contains ordered pairs that have both negative - and
-coordinates.
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The point on the coordinate plane with coordinates lies _____
On the coordinate plane, a point with a negative -coordinate and a positive
-coordinate lies in the upper left quadrant - Quadrant II.
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Which of the following could be a value of for
?
The graph is a down-opening parabola with a maximum of . Therefore, there are no y values greater than this for this function.
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What is the domain of ?
The domain of the function specifies the values that can take. Here,
is defined for every value of
, so the domain is all real numbers.
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Define
What is the natural domain of ?
The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which
27 is the only number excluded from the domain.
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Which of the following graphs does NOT represent a function?
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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Suppose .
To obtain the graph of , shift the graph
a distance of
units .
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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The figure above shows the graph of y = f(x). Which of the following is the graph of y = |f(x)|?
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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Below is the graph of the function :
Which of the following could be the equation for ?
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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What is the domain of ?
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
.
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What is the domain of the function ?
To find the domain, we must find the interval on which is defined. We know that the expression under the radical must be positive or 0, so
is defined when
. This occurs when
and
. In interval notation, the domain is
.
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Define the functions and
as follows:
What is the domain of the function ?
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,
, or
Since the domains of and
are the same, the domain of
is also the same. In interval form the domain of
is
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Define .
What is the natural domain of ?
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:
The domain is the set of all real numbers except those two - that is,
.
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True or false: The graph of has as a horizontal asymptote the graph of the equation
.
is a rational function whose denominator polynomial has degree greater than that of its numerator polynomial (2 and 1, respectively). The graph of such a function has as its horizontal asymptote the line of the equation
.
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Define
What is the natural domain of ?
Since the expression is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which
. We solve for
by factoring the polynomial, which we can do as follows:
Replacing the question marks with integers whose product is and whose sum is 3:
Therefore, the domain excludes these two values of .
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What is the equation for the line pictured above?
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,
which yields .
Therefore the equation of the line becomes:
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The chord of a central angle of a circle with circumference
has what length?
A circle with circumference has as its radius
.
The circle, the central angle, and the chord are shown below:
By way of the Isosceles Triangle Theorem, can be proved equilateral, so
, the correct response.
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The chord of a central angle of a circle with area
has what length?
The radius of a circle with area
can be found as follows:
The circle, the central angle, and the chord are shown below:
By way of the Isosceles Triangle Theorem, can be proved equilateral, so
, the correct response.
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