PLS help y=-x^2+2x graph
the coefficient of the parabola is -1 so it will be opening downwards... it's x-intercepts are x=0 and x=-2. which means the vertex is somewhere where x=-1. can you figure out the y-coordinate of the vertex?
how did you get -2
factor the equation: y = -x^2 + 2 = -x*(x - 2) oops. i mean x=2.. sorry.
still having trouble sorry
uh hello?
sorrry still here
do you understand the factoring?
so the vertex lies between these two x-intercepts at x=1...
sorry for being late but i am confused now because your way is different then what the book says
ok... i tell you what... i'll explain it the way I do it and you can see that we arrive at the same answer...
ok then
there is more than one way to do this problem....
wait doesn't x=-1?
forget that... that was before you found my mistake. the vertex now should be in the middle of x=0 and x=2. so x=1 is the x-coordinate of the vertex.
are you clear up to here?
uh...still confused alittle....
y = -x^2 + 2x is your equation. i factored the expression so I could find where the graph crosses the x-axis: y = -x(x-2) finding where the graph crosses the x-axis means you replace y with 0: 0 = -x(x-2) obviously, x=0, x=2.
The way that my book does you find the axis symmetry with x=-b/2a
i think....
so you want to do it the way your book does it then?
yah if it is ok with you
ok, identify these for me: a = b = c =
forget c.
a=-1 b=2 c=0
so x=-b/(2a) = -2/(2*(-1)) =
OH!!! i know what i did wrong :P
:)
but could you still continue?
yes... what did we just find? or what is the significance of x=1?
Huh??
you just calculated x = -b/(2a)... what's that for? don't say because the book does it..
is the vertex (1,1)?
yes...
but how do you find the other four points?
what other four? your problem just said to graph it...
oh. i see. you want verification points...
It says use the vertex and at least 4 other points
fill out this table for me: |dw:1334802529894:dw|
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