Question 718316
First of all, please put fractions which are factors in parentheses. The right side, as you posted it, means:
{{{-2/3x+2}}}
It was only your mention that the y-intercept was 2 that allowed me to figure out that it was supposed to be:
{{{(-2/3)x+2}}}<br>
description that the <br>At the y-intercept of 2, the y-coordinate is 2. The x-coordinate on the y-axis is always 0. This explains the x=0 solution.<br>
As for the other solution...
The graphs of equations of the form:
y = | mx+b |
will "bounce" off the x-axis. They will "bounce" at the point whose x-coordinate is the number that makes mx+b be zero. For y = | 2x+2 |, the graph will "bounce" off the x-axis at -1 because -1 makes 2x+2 be zero. (If you can't see this, just set 2x+2 = 0 and solve for x.)<br>
The graph of y = | mx + b | will be the graphs of two "half-lines" which meet at the "bounce" point. One "half-line" will the part of the graph of y = mx+b that is above the x-axis and the other "half-line" will the the part of the graph of y = -(mx+b) (or y - -mx-b) which is above the x-axis.<br>
For y = | 2x+2 |, the two "half-lines" would be: y = 2x+2 and y = -2x-2. The first "half-line" is the one whose y-intercept is 2 and intersects {{{y = (-2/3)x + 2}}} there. The other "half-line", y = -2x-2, will intersect {{{y = (-2/3)x + 2}}} somewhere to the left of the "bounce" point at (-1, 0).<br>
Here's a graph of both y = | 2x+2 | and {{{y = (-2/3)x + 2}}}:
{{{graph(400, 400, -5, 5, -2, 8, abs(2x+2), (-2/3)x + 2)}}}