Which of the following is the correct equation for this function? shows a parabola graph, vertex- (1,-2) x-inter: 0 and 2 A. y=2x^2-4x B. y+1=2(x-3)^2 C. y=(x+1)(x-2) D. y=2x^2+x-2
ok so xintercepts means that y is 0 there right? so you have f(0)=0 and f(2)=0 so the vertex form of a parabola is f(x)=a(x-h)^2+k but (h,k)=(1,-2)
so we have f(x)=a(x-1)^2-2 we can find a by using f(0)=0
but i don't know if it will also work for f(2)=0 we can try it once we get the equation
so what is f(0)? plug in 0 whereever you see an x
another danged over-determined parabola
lol
like the 4th one i have seen in a week. only need vertex and one other point.
but you know what zarkon said something once i think the vertex=1/2 (the x intercept + the other x-intercept)
who is making up these problems? you have the vertex is (h,k) then the parabola is \[y=a(x-h)^2+k\] and you are only permitted one other point to find a. everything else is fixed
or you know the h part of the vertex (the x-coordinate)
sure but so what? you have the vertex is (1,-2) so you know you have \[y=a(x+1)^2-2\]
right!
y=a(x-1)^2-2
one other point does it. two others is either redundant or impossible
the only thing is that you need a minue satellite between x and 1
yeah what you said
minus*
these questions irk me. like saying find the equation of the line between (1,3) , (2,5) and (3,7) either two points are sufficient, or there isn't one
lol
i just need to kno the equation
ok i got you started kassia but love for you to try to finish it with my help
we have f(x)=a(x-1)^2-2 use f(0)=0 to find a
damn i wrote it all out
it is answer 1
\[f(x)=a(x-1)^2-2\] \[f(0)=0=a(0-1)^2-2\] \[0=a-2\] \[a=2\] \[f(x)=2(x-1)^2-2\] \[f(x)=2(x^2-2x+1)-2\] \[f(x)=2x^2-4x\]
thank you
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