Question

Vertex (0,1) x intercepts is - 1 and 1 what is the equation of the parabola?

Answer 1

`\bf y=a(x-{{ h}})^2+{{ k}}\\ x=a(y-{{ k}})^2+{{ h}}\qquad\qquad vertex\ ({{ h}},{{ k}})`

those are the vertex form of a parabola... so hmmm
the vertex of this one is at 0,1 and intercepts or "solutions" are at -1 and 1, so is opening downwards, notice the picture below

that means, the squared variable is the "x", thus the form is
`\bf y=a(x-{{ h}})^2+{{ k}}`

now, we know the vertex is at 0,1, and two x-intercepts of
`\pm 1,0`

thus
`\bf y=a(x-{{ h}})^2+{{ k}}\qquad \begin{cases} h=0\\ k=1\\ when \\ x=\pm 1\\ y=0 \end{cases} \\\\\\ \textit{so.. hmmm let us use the point ... say hmmm 1,0} \\\\ y=a(x-{{ h}})^2+{{ k}}\implies 0=a(1-0)^2+1`

solve for "a", to see what that coefficient is, then plug it back in the vertex form equation