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

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