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A parabola $ax^2+bx+c$ contains the points $(-1,0)$, $(0,5)$, and $(5,0)$. Find the value $100a+10b+c$.

 Nov 10, 2019
 #1
avatar+2482 
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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?

 Nov 10, 2019

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