Graph the inequality -2x - y - 1
inequality?
yes thats what it says
this is what I see => \(\huge -2x - y - 1 \)
Yes i can even take a picture of my wkst...
hmmm. ok... I take a picture then
because is missing an inequality sign
ok wait like few min
it's #7
hmm, is a typo then, if I were to make anything out of that, I'd think is \(\huge -2x\ge -y-1\ or \ -2x \le -y-1\)
alright then Ill skip it.. can you help me with #6 then
well, you'd solve for "y" first what would that give you?
so divide by -2?
yes, keep in mind that, when multiplying/dividing/exponentiazling by a negative value, you need to \(\bf \text{flip}\) the sign
\[y \ge x/2 + 1\]
yeap, so in essence the graph will be, using the line of y = x/2 + 1 the line drawn will be SOLID, because is using \(\bf \ge\) for > or < is a DASHED line
is a straight line, so all you need is 2 points
how do i graph it? the x/2 part
well, one of the easiest way to get 2 points, is set x = 0, then set y = 0 so \(\bf y \ge \cfrac{x}{2}+1\\ \quad \\ \quad \\ x =0,\qquad y = \cfrac{0}{2}+1\implies y =1\quad (0, 1)\\ \quad \\ y=0, \qquad 0 = \cfrac{x}{2}+1 \implies -2 = x\quad (-2, 0)\)
|dw:1379979450237:dw|
do i need to find the points or can i just look at the equation?
yeap, that's the line so for the shading, we do some testing on a point NOT IN THE LINE say let's use (0, 0) is not in the line, so we'll use that \(\bf y \ge \cfrac{x}{2}+1\\ \quad \\ \quad \\ (0,0)\\ \quad \\ y \ge \cfrac{x}{2}+1\implies 0 \ge \cfrac{0}{2}+1\implies 0\ge 1\) so, is 0 really greater or equal to 1? well, not
so the region where (0,0) is at, is the FALSE area so the TRUE area will be on the other side of the line, and that's the one that gets the shading
do i need to find the points or can i just look at the equation?
for the shading?
no to graph it
is x/2 the same as 2x?
hmm, no.... x/2 is one half, 2x is twice as much
to graph it? you just did :)
the graph is just that one line you drew
then why did you find the (0,1) and (-2,0)
ohh that's for the shading to show in the graph, where the inequality applies that is, what part of the graph the values for which that inequality http://www.regentsprep.org/regents/math/algebra/ae85/grineqa.htm
hhhmmm shoot I misunderstood.. anyhow... the ((0,1) and (-2,0) part was just to get 2 points to make the line, but you did just fine
the (0,0) testing part is for the shading of it
ohh so my graph is okay too?
yeap
alright thanks!! would you mind helping me with one more thing?
@JuanitaM
@jdoe0001
Join our real-time social learning platform and learn together with your friends!