Question

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

Answer 1

`100a+10b+c=-55`

Explanation:

Solve for c for each coordinate point:

`f(x)=ax^2+bx+c`

`0=ax^2+bx+c`

`0=a(-1)^2+b(-1)+c`

`0=a-b+c`

`-c=a-b`

`c=b-a`

`f(x)=ax^2+bx+c`

`5=a(0)^2+b(0)+c`

`5=c`

`f(x)=ax^2+bx+c`

`0=a(5)^2+b(5)+c`

`0=25a+5b+c`

`-c=25a+5b`

`c=-25a-5b`

Combine first and third equations to solve for b:

`b-a=-25a-5b`

`6b-a=-25a`

`6b=-24a`

`b=-4a`

Substitute values of b=-4a and c=5 to solve for a:

`c=b-a`

`5=-4a-a`

`5=-5a`

`-1=a`

Find b:

`c=b-a`

`5=b-(-1)`

`5=b+1`

`4=b`

Evaluate required expression:

`100a+10b+c`

`100(-1)+10(4)+5`

`-100+40+5`

`-60+5`

`-55`

Therefore,
`100a+10b+c=-55`