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I know how to find a parabola passing through $(0, 0)$ and $ (2, 0)$:

$y=x(x-2)$

I also know how to find a parabola with vertex at $(1, −2)$: $(x-1)^2-2$

But I don't know how to find a parabola with both vertex at $(1, −2)$ passing through $(0, 0)$ and $(2, 0)$?

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  • $\begingroup$ The first is part way there. You forgot the coefficient out front. It should be $ax(x-1)$ $\endgroup$ – Cameron Williams Sep 9 '19 at 10:50
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A parabola passing through $\left(0,0\right)$ and $\left(2,0\right)$ has also the form $y=ax(x-2)$, then you can put in $\left(-2,1\right)$ to find $a$.

Also, you can take the form $y=c(x-1)^2-2$ and plug in $\left(0,0\right)$

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Just go step by step! A parabola is determind by three unkowns $a,b,c$, with a expression $f(x)=ax^2+bx+c$

You know three points on the parabola, so: $$f(0)=0$$ $$f(2)=0$$ $$f(1)=-2$$

Now you can get values for $a,b$ and $c$. Then make sure that $(1, -2)$ is indeed the vertex and not just a random point in the parabola.

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You've already come some way as you know what to do with the given $x$ intercepts of $0$ and $2$. So I'll continue from where you left off. Basically the form is $y =kx(x-2)$, where $k$ is a constant you have to find.

Note that when $x=1,y=-2$ which means $-2 = k(1)(1-2)$ giving $k=2$ and the answer of $y =2x(x-2)$.

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