Math  /  Algebra

Question4) 5x2y+8=05 x-2 y+8=0 Rewrite in slope-intercept form. 5) State the domain and range of the function y=2x+8y=2 x+8
Domain: Range: \qquad

Studdy Solution

STEP 1

What is this asking? We need to rewrite an equation in a different form and then figure out the possible input and output values for another equation. Watch out! Don't mix up domain and range!
Also, remember slope-intercept form is y=mx+by = mx + b, not the other way around!

STEP 2

1. Rewrite in slope-intercept form
2. Find the domain
3. Find the range

STEP 3

We want to **isolate** the yy term to get it by itself.
Our equation is 5x2y+8=05x - 2y + 8 = 0.
Let's **subtract** 5x5x and 88 from both sides to move everything away from the yy term.
This gives us 2y=5x8-2y = -5x - 8.
Remember, what we do to one side, we *must* do to the other!

STEP 4

Now, we **divide** both sides by 2-2 to get yy all alone.
So, we have 2y2=5x82 \frac{-2y}{-2} = \frac{-5x - 8}{-2} which simplifies to y=5x2+82 y = \frac{-5x}{-2} + \frac{-8}{-2} giving us our **slope-intercept form**: y=52x+4 y = \frac{5}{2}x + 4

STEP 5

The **domain** is all the possible xx values we can plug into our equation y=2x+8y = 2x + 8.

STEP 6

Can we plug in any number for xx?
Yes! There are no fractions with xx in the denominator or square roots to worry about.
So, xx can be any real number.
We write that as (,) (-\infty, \infty) .

STEP 7

The **range** is all the possible yy values that come out of our equation y=2x+8y = 2x + 8.

STEP 8

Since the domain is all real numbers, and our equation is a line, the range is also all real numbers!
The line goes up and down forever.
So, the range is also (,) (-\infty, \infty) .

STEP 9

The equation in slope-intercept form is y=52x+4y = \frac{5}{2}x + 4.
The domain of y=2x+8y = 2x + 8 is (,)(-\infty, \infty) and the range is (,)(-\infty, \infty).

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