161,081 views
45 votes
45 votes
Determine whether the relation y = -13x- 7 defines y as a function of x. Give the domain.Is the equation a function?A. Yes, the equation y=- 13x- 7 assigns exactly one y-value to each distinct x-value in the domain.B. Yes, because the equation y = - 13x- 7 assigns at least one y-value to each distinct x-value.C. No, because the equation y = - 13x- 7 assigns more than one y-value to one of the x-values.D. No, because the equation y = - 13x- 7 assigns exactly one x-value to each distinct y-value.The domain is _______

User Paul Freez
by
2.1k points

1 Answer

7 votes
7 votes

We have the following relation:


y=-13x-7

And we have to determine if that relationship between both variables is a function.

1. To achieve that, we need to remember that:

• A function is a relation in which there is exactly one output for each input.

,

• The set of all the inputs for a function is the domain of the function.

,

• The set of all the outputs for a function represents the range of the function

,

• If we have to define a function, we have to describe the rule of that function.

2. We can see that the relation above is of the form:


y=mx+b

Where

• m is the slope of the line. In this case, m = -13, which is a negative slope.

,

• b is the y-intercept of the line, and it is the point where the line passes through the y-axis, and when x = 0. In this case, b = -7.

Then the rule of the function is given by y = -13x - 7.

3. Therefore, this is a linear function, and in this case, we have that the rule assigns exactly one y-value to each distinct x-value in the domain. Likewise, the domain in this kind of function is the set of all the real numbers, and we can write it in interval notation as follows:


\text{ Domain=\lparen-}\infty,\infty)

Hence, in summary, we can say that:

A. Yes, the equation y =- 13x- 7 assigns exactly one y-value to each distinct x-value in the domain.

The domain is (in interval notation) is:


\begin{equation*} \text{ Domain=\lparen-}\infty,\infty) \end{equation*}

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.