# A parabola with vertex at $(1, −2)$ passing through $(0, 0)$ and $(2, 0)$

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 $$(2, 0)$$?

• The first is part way there. You forgot the coefficient out front. It should be $ax(x-1)$ – Cameron Williams Sep 9 '19 at 10:50

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 $$\left(0,0\right)$$

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.

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 $$y =2x(x-2)$$.