Let’s look at the first equation, y = 3 |1-3x| + 2.

You can think of the function 3|1-3x|+2 as having two pieces, one where the absolute-value is leaving positive numbers alone, and one where it’s making negative numbers positive.

First figure out what values of x make (1-3x) negative, and what values of x don’t.

Think of solving 1-3x=0; you’ll get x=1/3. Does 1-3x go negative when x goes below 1/3 or above 1/3? Why?

We can rewrite the equation as

y = 3(1-3x) + 2 if x <= 1/3 y = 3(3x-1) + 2 if x > 1/3

(3x-1) is the same as (-(1-3x)); representing what |1-3x| does to negative numbers inside.

Simplifying,

y = 5 – 9x if x <= 1/3 y = 9x - 1 if x > 1/3

The second problem, y = -2|x+7|-4, also breaks down into two pieces, one where the absolute-value’s result is (x+7) “as is”, one where it changes negative to positive, as -(x+7).

A slightly more roundabout way of doing this is to get the absolute values by themselves and make two equations like a=b and a=-b (from a problem like IaI=b). Solve and graph them both and you should get 2 lines (and 2 V’s.). Test a point on one of the lines (above or below the intersection). If the equation is true, erase the half of the V that you didn’t test. If it’s false, erase the half of the V that you did. (the V should go up or down, not sideways).

On 1, for example, the v point is when x = 1/3. y = 2 there. So try 0 and -1 for x, then 1 and 2.

1) y = 3 |1 – 3x| + 2

The domain is ( -∞, +∞), x could have any value.

For any value of x, y is positif (absolute value). and y is always >= 2 .

That’s mean the range is [2, +∞).

Try to illustrate and graph the function f = |1 – 3x|

Then graph 3f = 3|1 – 3x| that’s mean stretch the function 3 times vertically.

Then, graph 3f + 2 = y = 3|1 – 3x| + 2. that’s mean increase the function 3f, 2 units on the y-axis.

And that’s it.

Try to think about the second one the same way.

Good luck