Coordinate Geometry - Geometry
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?
In the standard xy coordinate plane, what quadrant(s) could point A lie in, if its x-coordinate and y-coordinate have opposite signs?
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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. (+, -).
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. (+, -).
The point on the coordinate plane with coordinates 
lies
The point on the coordinate plane with coordinates
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On the coordinate plane, a point with a negative 
-coordinate and a negative 
-coordinate lies in the lower left quadrant - Quadrant III.
On the coordinate plane, a point with a negative
Which of the following ordered pairs is in Quadrant III?
Which of the following ordered pairs is in Quadrant III?
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By definition, only Quadrant III contains ordered pairs that have both negative 
- and 
-coordinates.
By definition, only Quadrant III contains ordered pairs that have both negative
The point on the coordinate plane with coordinates 
lies
The point on the coordinate plane with coordinates
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On the coordinate plane, a point with a negative 
-coordinate and a positive 
-coordinate lies in the upper left quadrant - Quadrant II.
On the coordinate plane, a point with a negative
Which of the following could be a value of f(x) for $f(x)=-x^2$ + 3?
Which of the following could be a value of f(x) for $f(x)=-x^2$ + 3?
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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.
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.
What is the domain of y = 4 - $x^{2}$?
What is the domain of y = 4 - $x^{2}$?
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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.
The domain of the function specifies the values that
Define 
What is the natural domain of 
?
Define
What is the natural domain of
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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.
The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression
27 is the only number excluded from the domain.
Which of the following graphs does NOT represent a function?
Which of the following graphs does NOT represent a function?
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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.
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.
Suppose 
.
To obtain the graph of 
, shift the graph 
a distance of 
units .
Suppose
To obtain the graph of
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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.
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.
The figure above shows the graph of y = f(x). Which of the following is the graph of y = |f(x)|?
The figure above shows the graph of y = f(x). Which of the following is the graph of y = |f(x)|?
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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.
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.
Below is the graph of the function 
:
Which of the following could be the equation for 
?
Below is the graph of the function
Which of the following could be the equation for
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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.
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.
What is the domain of y=-2$\sqrt{x}$?
What is the domain of y=-2$\sqrt{x}$?
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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 xgeq 0.
To find the domain, we need to decide which values
What is the domain of the function $y=$\sqrt{4-x^{2}$$}?
What is the domain of the function $y=$\sqrt{4-x^{2}$$}?
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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 
.
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
Define the functions 
and 
as follows:
What is the domain of the function 
?
Define the functions
What is the domain of the function
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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 
The domain of
Since the domains of
Define 
.
What is the natural domain of 
?
Define
What is the natural domain of
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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,
.
The only restriction on the domain of
The domain is the set of all real numbers except those two - that is,
True or false: The graph of 
has as a horizontal asymptote the graph of the equation 
.
True or false: The graph of
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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 
.
Define 
What is the natural domain of 
?
Define
What is the natural domain of
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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 
.
Since the expression
Replacing the question marks with integers whose product is
Therefore, the domain excludes these two values of
What is the equation for the line pictured above?
What is the equation for the line pictured above?
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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:
A line has the equation
The
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:
The chord of a 
central angle of a circle with circumference 
has what length?
The chord of a
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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.
A circle with circumference
The circle, the central angle, and the chord are shown below:
By way of the Isosceles Triangle Theorem,
The chord of a 
central angle of a circle with area 
has what length?
The chord of a
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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.
The radius
The circle, the central angle, and the chord are shown below:
By way of the Isosceles Triangle Theorem,