Question

Find the quadratic equation using the points given (-1,2), (0,1) and (-2,5).

Answer 1

The general equation for a quadratic equation is,

`y=ax^2+bx+c`

Substititute the values to obtain the equations for the coefficients.

`\begin{gathered} 2=a(-1)^2+(-1)b+c \\ a-b+c=2 \end{gathered}`
`\begin{gathered} 1=a(0)^2+b(0)+c \\ c=1 \end{gathered}`

and

`\begin{gathered} 5=a(-2)^2+b(-2)+c \\ 4a-2b+c=5 \end{gathered}`

Substitute the value of c in the equation a-b+c=2 to obtain the equation for a and b.

`\begin{gathered} a-b+1=2 \\ a=1+b \end{gathered}`

Substitute the value of a and c in the equation 4a-2b+c=5 to obtain the value of b.

`\begin{gathered} 4(1+b)-2b+1=5 \\ 4-2b+1=5 \\ 2b=0 \\ b=0 \end{gathered}`

Substitute the value of b in the equation a=1+b to obtain the value of a.

`\begin{gathered} a=1+0 \\ a=1 \end{gathered}`

So quadratic equation for a=1, b=0 and c=1 is,

`y=x^2+1`