How to graph a function - Geometry

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 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.

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.

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.

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.

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.

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 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 .

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 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,

.

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 .

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 .

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 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.

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.

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 .

The radius of a circle with area can be found as follows:

The circle, the central angle, and the chord are shown below, along with , which bisects isosceles

We concentrate on , a 30-60-90 triangle. By the 30-60-90 Theorem,

and

The chord has length twice this, or

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 .

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 .