SOLUTION: Match the graph with one of the equations. Crosses at (0,2) & (2,0) A) y = –x – 2 B) y = x – 2 C) y = –x + 2 D) y = x + 2

Algebra ->  Algebra  -> Systems-of-equations -> SOLUTION: Match the graph with one of the equations. Crosses at (0,2) & (2,0) A) y = –x – 2 B) y = x – 2 C) y = –x + 2 D) y = x + 2       Log On

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Question 105459: Match the graph with one of the equations.
Crosses at (0,2) & (2,0)

A) y = –x – 2
B) y = x – 2
C) y = –x + 2
D) y = x + 2

Answer by MathLover1(3376) About Me  (Show Source):
You can put this solution on YOUR website!
A y+= -x – 2
B y = x – 2
C y+= -x + 2
D y+=+x+%2B+2
ans.:
C y+= -x + 2
if x=0 => y+=+2
and, if y=0 => x+=+2


Solved by pluggable solver: Graphing Linear Equations
In order to graph y=-1%2Ax%2B2 we only need to plug in two points to draw the line

So lets plug in some points

Plug in x=-7

y=-1%2A%28-7%29%2B2

y=7%2B2 Multiply

y=9 Add

So here's one point (-7,9)


drawing%28+600%2C+600%2C+-17%2C4%2C+-2%2C+19%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-7%2C9%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-7%2C9%2C+.1%2C+1.5+%29+%29%0D%0A++%29

Now lets find another point

Plug in x=-6

y=-1%2A%28-6%29%2B2

y=6%2B2 Multiply

y=8 Add

So here's another point (-6,8). Add this to our graph


drawing%28+600%2C+600%2C+-17%2C4%2C+-2%2C+19%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-6%2C8%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-6%2C8%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-7%2C9%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-7%2C9%2C+.1%2C+1.5+%29+%29%0D%0A++%29


Now draw a line through these points

drawing%28+600%2C+600%2C+-17%2C4%2C+-2%2C+19%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-17-10%2C-1%2A%28-17-10%29%2B2%2C4%2B10%2C-1%2A%284%2B10%29%2B2%29%2C%0D%0A++++blue%28+circle%28+-6%2C8%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-6%2C8%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-7%2C9%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-7%2C9%2C+.1%2C+1.5+%29+%29%0D%0A++%29 So this is the graph of y=-1%2Ax%2B2 through the points (-7,9) and (-6,8)


So from the graph we can see that the slope is -1%2F1 (which tells us that in order to go from point to point we have to start at one point and go down -1 units and to the right 1 units to get to the next point), the y-intercept is (0,2)and the x-intercept is (2,0)


We could graph this equation another way. Since b=2 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,2).


So we have one point (0,2)


drawing%28+600%2C+600%2C+-10%2C11%2C+-9%2C+12%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++%0D%0A++++blue%28+circle%28+0%2C2%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+0%2C2%2C+0.14-0.05%29+%29%0D%0A++%29


Now since the slope is -1%2F1, this means that in order to go from point to point we can use the slope to do so. So starting at (0,2), we can go down 1 units


drawing%28++600%2C+600%2C+-10%2C11%2C+-9%2C+12%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+0%2C2%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+0%2C2%2C+0.14-0.05%29+%29%0D%0A++blue%28+arc%28+0%2C+2%2B%28-1%2F2%29%2C+2%2C+-1%2C+90%2C+270+%29+%29%0D%0A++%0D%0A++%29
and to the right 1 units to get to our next point

drawing%28++600%2C+600%2C+-10%2C11%2C+-9%2C+12%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+1%2C1%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+1%2C1%2C+0.14-0.05+%29+%29%2C%0D%0A++++blue%28+circle%28+0%2C2%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+0%2C2%2C+0.14-0.05%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+2%2B%28-1%2F2%29%2C+2%2C+-1%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%281%2F2%29%2C+1%2C+1%2C+2%2C+0%2C+180+%29+%29%0D%0A++%0D%0A++%29
Now draw a line through those points to graph y=-1%2Ax%2B2


drawing%28+600%2C+600%2C+-10%2C11%2C+-9%2C+12%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-10-10%2C-1%2A%28-10-10%29%2B2%2C11%2B10%2C-1%2A%2811%2B10%29%2B2%29%2C%0D%0A++++blue%28+circle%28+1%2C1%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+1%2C1%2C+0.14-0.05+%29+%29%2C%0D%0A++++blue%28+circle%28+0%2C2%2C+0.14%29%29%2C%0D%0A++++blue%28+circle%28+0%2C2%2C+0.14-0.05%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+2%2B%28-1%2F2%29%2C+2%2C+-1%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%281%2F2%29%2C+1%2C+1%2C+2%2C+0%2C+180+%29+%29%0D%0A++%29 So this is the graph of y=-1%2Ax%2B2 through the points (0,2) and (1,1)