What makes |x-7| x >/= 4 turn upwards?

Question

Hi,

 

I have the problem -1/3x-4, x<3  |x-7| x > 4. When my teacher graphed the second function, the x-intercept was at 7, with y-intercept 0, of course, but it went in a v-shape at an x slope and reached (4,4), and continued at the same slope in the other direction. While I understand it stopping at x=4, why does it also stop at y=4? Also, wouldn't the line go below y=0 as well until x=4?

 

Thanks,

-Gabe

Answer 1

You apparently have a PIECEWISE function definition:

when x<3, f(x) = (-1/3)x-4 ... this piece is a straight line having negative one third as its slope, and has y intercept -4;

when x>4, there's an absolute value definition.

 

Apparently function f is not defined on [3,4]. The V-shape terminates short of x=4 BECAUSE OF THE INEQUALITY DEFINING THE domain. It is improper to say "it went at an x slope..."; just say that the slope of the V-shape is -1, when x<7, and it's +1 to the right of x=7.  I don't see any "stop at y=4"  (If you're referring to the piece of the graph defined by an absolute value, you're referring to the second PIECE, not second function.)