+18 A parabola $ax^2+bx+c$ contains the points $(-1,0)$, $(0,5)$, and $(5,0)$. Find the value $100a+10b+c$.
+2863 Plug in the x values in the coordinates into the parabola equation. then set it equal to the y-value in the coordinates.
So we get 3 equations
\(a(-1)^2+b(-1)+c=0\)
\(a(0)^2+b(0)+c=5\)
\(a(5)^2+b(5)+c=0\)
Simplifies to:
\(a-b+c=0\)
\(c=5\)
\(25a+5b+c=0\)
Plugging C we get two equations
\(a-b+5=0\)
\(25a+5b+5=0\)
We simplify:
\(a-b=-5\)
\(25a+5b=-5\)
We simplify further:
\(a-b=-5\)
\(5a+b=-1\)
We use elimination (addition):
\(6a=-6\)
\(a=-1\)
Plugging back a = -1 into \(a-b=-5\)
We get \(b=4\)
So we have \((a,b,c)=(-1,4,5)\)
I will leave evaluating \(100a+10b+c\) up to you now.
Can you solve the problem?
