Question

There is a unique parabola whose symmetry axis is parallel to the y-axis, and that passes through the three points (1, 1), (−2, −2), and (0, −4). Write an equation for it. Given any three points, must there be a parabola that will pass through them? Explain.

Answer 1

since the parabola has an axis of symmetry parallel to the y-axis, then we know it's a vertical parabola.

`y=ax^2+bx+c~\hfill (1~~,~~1)\hspace{5em}(-2~~,~-2)\hspace{5em}(0~~,~-4) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} x=1\\ y=1 \end{cases}\implies 1=a(1)^2+b(1)+c\implies \hspace{6em}1=a+b+c \\\\\\ \begin{cases} x=-2\\ y=-2 \end{cases}\implies -2=a(-2)^2+b(-2)+c\implies\qquad -2=4a-2b+c \\\\\\ \begin{cases} x=0\\ y=-4 \end{cases}-4=a(0)^2+b(0)+c\implies \hspace{6em}\boxed{-4=c}`

now le'ts notice, we ended up wtih three equations on the far-right, and low and behold, -4=c, hell we can just plug that in the other two and only have two equations after all, most times we'd have three btw, this time we can boil it down to two, and let's use substitution for them.

`1=a+b+c\implies 1=a+b-4\implies 5=a+b\implies 5-b=a \\\\[-0.35em] ~\dotfill`

`-2=4a-2b+c\implies -2=4a-2b-4\implies 2=4a-2b\implies 1=\cfrac{4a-2b}{2} \\\\\\ 1=2a-b\implies \stackrel{\textit{substituting from the 1st equation}}{1=2(5-b)-b}\implies 1=10-2b-b \\\\\\ 1=10-3b\implies -9=-3b\implies \cfrac{-9}{-3}=b\implies \boxed{3=b} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that}}{5-b=a}\implies 5-3=a\implies \boxed{2=a} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\Large \begin{array}{llll} y=2x^2+3x-4 \end{array}} ~\hfill`

Check the picture below.