Question

Please answer these:)

Answer 1

Explanation:

Not a function

In order for an equation to represent a function any single value of x must have at most one corresponding value of y which satisfies the equation.

For

x2+y2=9

(for example)

x=0

there are two values for y (namely +3 and −3) which satisfy the equation

If you plug this graph into a calculator, you will find that it is a graph of a circle.

Circles have the formula x2+y2=r2 if they are centered at the origin.

From that, we can find that the radius is 3, and it is centered at the origin.

To find the domain and range, we just find the largest and smallest x and y values.

Since the radius is 3, and the graph is centered at the origin, the largest x value is 3. Similarly, the smallest x value is -3.

The domain is −3≤x≤3

We can use the same thing for the range. The radius is 3, so the largest and smallest values of the y will be 3 and -3.

The range is -3≤y≤3

Answer 2

See below

Explanation:

6. Look at the graph of the equation x^2+y^2=8. Give its domain and range.

• Both domain and range are: (- √8, √8) or (-2.828, 2.828)

7. Use algebraic means to show that x^2+y^2=8 is not a function. Explain your process.

• The vertical line test. If more than one point is crossed on the graph by a vertical line, then the relation is not considered to be a function.

8. Is there any value(s) of the domain of x^2+y^2=8 that passes the vertical line test? If so, name the value(s) and state whether or not the existence of this value makes this relation a function. You can use Desmos to help you explore this idea, if needed.

• There are two points - 2.828 and 2.828 that pass the vertical line test but all the other points don't, therefore this relation is not a function