Question

30 POINTS!!!!!!!!PLS COME AND LOOK!!! I NEED YOU GENIUSES!!! I WILL GIVE BRAINLIEST!!!!! AT LEAST COME AND LOOK!!!! WILL FOREVER BE GREAT FULL!!! EASY IM JUST DUMB!! The graph represents the equation 2 + 2 = 8. Use this graph to answer questions 6, 7, and 8. PART A) Look at the graph of the equation 2 + 2 = 8. Give its domain and range. PART B) Use algebraic means to show that 2 + 2 = 8 is not a function. Explain your process.

Answer 1

Answer:

See below.

Step-by-step explanation:

So we have the graph of:

$x^2+y^2=8$

As given, the graph is a circle with a radius of 2√2 and a center of (0,0)

Part 1)

The domain of a function is the span of x-values it covers and the range of a function is the span of y-values it covers.

From the graph, we can see that the x-values the graph covers goes from -2√2  to +2√2. The graph includes these values.

Thus, the domain is, in interval notation:

$[-2\sqrt2,2\sqrt2]$

From the graph, we can see that the y-values the graph covers also goes from -2√2 to 2√2.

Therefore, like the domain, the range in interval notation is:

$[-2\sqrt2,2\sqrt2]$

Part 2:

First of all, we can use the vertical line test to prove that the graph is indeed not a function.

Algebraically, we have to do some more.

First, recall what it means when an equation is not a function. It means that one x maps onto two or more ys.

In other words, to prove that the equation is not a function, we can substitute an x for a value and solve for y.

For instance, let's use 0 for x (and 0 is within the domain). Thus:

$x^2+y^2=8\\(0)^2+y^2=8$

Simplify:

$y^2=8$

Take the square root of both sides:

$y=\pm\sqrt8\\y=\pm2\sqrt2$

In other words, when x is 0, y can be either -2√2 or 2√2. So we have two different y-values with the same x-coordinate.

$(0,-2\sqrt2)(0,2\sqrt2)$

Therefore, the equation is not a function.