91Ó°ÊÓ

The points:

$(0,1),(1,0),(-2,1)$

A quadratic function:

$a{x}^2+bx+c=0$

For the point to pass through the function, the point must satisfy the function.

So for $(0,1)$,

$$\begin{gathered} y=a{x}^2+bx+c \\ 1=a{\left(0\right)}^2+b\left(0\right)+c \\ c=1 \end{gathered}$$

Substitute the point $(1,0)$ in the equation, we get

$$\begin{gathered} 0=a{\left(1\right)}^2+b\left(1\right)+c \\ 0=a+b+c \end{gathered}$$

Substitute the point $(-2,1)$in the equation, we get

$$\begin{gathered} 1=a{(-2)}^2+b(-2)+c \\ 1=4a-2b+c \end{gathered}$$

The system of equations are $\left\{\begin{array}{l}c=1\\ a+b+c=0\\ 4a-2b+c=1\end{array}\right.$

By substituting $c=1$in the second and third equations, we get,

$$\begin{gathered} a+b+1=0 \\ a+b=-1 \\ 4a-2b+1=1 \\ 4a-2b=0 \end{gathered}$$

Since

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

Substitute $a=-1-b$in the above equation, we get,

$$\begin{gathered} 4a-2b=0 \\ 4(-1-b)-2b=0 \\ -6b-4=0 \\ -6b=4 \\ b=-\frac{2}{3} \end{gathered}$$

The value of $b$ is $-\frac{2}{3}$

Substitute $b=-\frac{2}{3};c=1$in the equation,

$$\begin{gathered} a+b+c=0 \\ a+(-\frac{2}{3})+1=0 \\ a+\frac{1}{3}=0 \\ a=-\frac{1}{3} \end{gathered}$$

The value of $a$ is $-\frac{1}{3}$

Substitute the values of a, b, c in the given equation, we get,

$$\begin{gathered} a{x}^2+bx+c=0 \\ -\frac{1}{3}{x}^2+(-\frac{2}{3}x)+1=0 \\ \frac{1}{3}{x}^2+\frac{2}{3}x-1=0 \end{gathered}$$