Question

The equation for a parabola has the form y=????x2+????x+cy=ax2+bx+c, where ????a, ????b, and cc are constants and ????≠0a≠0. Find an equation for the parabola that passes through the points (−1,9)(−1,9), (3,−19)(3,−19), and (−4,−12)(−4,−12)

Answer 1

The equation for this parabola satisfies

`\begin{cases}a-b+c=9\\9a+3b+c=-19\\16a-4b+c=-12\end{cases}`

(in other words, plug in each given point's coordinates
`(x,y)` into the equation
`y=ax^2+bx+c`)

Now,

`(a-b+c)-(9a+3b+c)=9-(-19)\implies-8a-4b=28\implies2a+b=-7`

`(a-b+c)-(16a-4b+c)=9-(-12)\implies-15a+3b=21\implies5a-b=-7`

and

`(2a+b)+(5a-b)=(-7)+(-7)\implies7a=-14\imples a=-2`

`2(-2)+b=-7\implies b=-3`

`(-2)-(-3)+c=9\implies c=8`

So the equation of the parabola is

`y=-2x^2-3x+8`