Functions and Graphs - High School Math

The equation of a circle is: where is the radius and is the center.

In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to , which is . We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 10.

To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:

Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:

and

We can then solve these two equations to obtain .

To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:

Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:

and

We can then solve these two equations to obtain

Use substution to solve this problem:

becomes and then is substituted into the second equation. Then solve for :

, so and to give the solution .

To solve for the -intercept, set to zero and solve for :

To solve for the -intercept, set to zero and solve for :

Slope-intercept form is .

Since the equation given in the question is based off of the parent function , we can write the general form for transformations like this:

determines the vertical stretch or compression factor.

• If is greater than 1, the function has been vertically stretched (expanded) by a factor of .
• If is between 0 and 1, the function has been vertically compressed by a factor of .

In this case, is 4, so the function has been vertically stretched by a factor of 4.

determines the horizontal stretch or compression factor.

• If is greater than 1, the function has been horizontally compressed by a factor of .
• If is between 0 and 1, the function has been horizontally stretched (expanded) by a factor of .

In this case, is 6, so the function has been horizontally compressed by a factor of 6. (Remember that horizontal stretch and compression are opposite of vertical stretch and compression!)

determines the horizontal translation.

• If is positive, the function was translated units right.
• If is negative, the function was translated units left.

In this case, is 3, so the function was translated 3 units right.

determines the vertical translation.

• If is positive, the function was translated units up.
• If is negative, the function was translated units down.

In this case, is -7, so the function was translated 7 units down.

To find the center or the radius of a circle, first put the equation in the standard form for a circle: , where is the radius and is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula .

, so .

and , so and .

Therefore, .

Because the constant, in this case 4, was not in the original equation, we need to add it to both sides:

Now we do the same for :

We can now find :

To find the center or the radius of a circle, first put the equation in standard form: , where is the radius and is the center.

From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula .

, so .

and , so and .

This gives .

Because the constant, in this case 9, was not in the original equation, we must add it to both sides:

Now we do the same for :

We can now find the center: (3, -9)

The domain includes the values that go into a function (the x-values) and the range are the values that come out (the or y-values). A sine curve represent a wave the repeats at a regular frequency. Based upon this graph, the maximum is equal to 1, while the minimum is equal to –1. The x-values span all real numbers, as there is no limit to the input fo a sine function. The domain of the function is all real numbers and the range is .

Using as the input () value for this equation generates an output () value that contradicts the stated condition of .

Therefore is not a valid value for and not in the equation's domain:

The domain of a function is the set of possible values for the variable. The range would be the possible values for the solution, .

The value inside of a square root must be greater than or equal to zero in order to have a real solution.

Now we can solve for in the inequality.

Square roots have positive and negative roots, so we need to set up two results. Remember to switch the direction fo the inequality for the negative solution.

These solutions can be combined to give our final answer.

Any values of that are not included will result in an imaginary (impossible) answer.

The domain of a function includes all of the values of x for which f(x) is real and defined. In other words, if we put a value of x into the function, and we get a result that isn't real or is undefined, then that value won't be in the domain.

If we let x = 0, then we will be forced to evaluate , which is equal to 1/0. The value of 1/0 is not defined, because we can never have zero in a denominator. Thus , because f(0) isn't defined, 0 cannot be in the domain of f(x).

The answer is 0.

The domain of a function consists of all of the possible values for x. In this case, we want to make sure that we are not dividing by ( in the denominator), since that would make our function undefined. Having would make the denominator . Thus, that is not in our domain. There is nothing else that would make this function undefined, and thus the domain is all real numbers except

Domain represents all possible values for . In this situation, we want to make sure that the numbers under the square root are greater than or equal to 0. This will be the case when . Thus, the domain includes all real numbers such that .

The range of a function is defined as the possible values for , or the possible outcomes. In this function, it is not possible to get any sort of negative number as an outcome. It is possible to get zero when . Thus, the range is all real numbers greater than or equal to 0.

The range of a function is all of the possible values that the equation can take. For this equation our value cannot be negative, as a negative number squared still gives us a positive value.

Since , we know that the lowest possible value that can reach is . Therefore the range is .

The vertical line test can be used to determine if an equation is a function. In order to be a function, there must only be one (or ) value for each value of . The vertical line test determines how many (or ) values are present for each value of . If a single vertical line passes through the graph of an equation more than once, it is not a function. If it passes through exactly once or not at all, then the equation is a function.

The horizontal line test can be used to determine if a function is one-to-one, that is, if only one value exists for each (or ) value. Calculating zeroes, domain, and range can be useful for graphing an equation, but they do not tell if it is a function.

Example of a function:

Example of an equation that is not a function:

This expression is the same as saying "take the answer of and plug it into ."

First, we need to find . We do this by plugging in for in .

Now we take this answer and plug it into .

We can find the value of by replacing with .

This is our final answer.