91Ó°ÊÓ

Find the quadratic function for which:

Contains the points (0, 5), (1, 2), and (3, 2).

The Quadratic function is in the form:

$f\left(x\right)=a{x}^2+bx+c$

Find the value of a, b, c.

The quadratic function contains the points so it satisfy the quadratic function.

Point (0,5) : $$\begin{gathered} f\left(0\right)=5 \\ a·0^2+b·0+c=5 \\ c=5 \end{gathered}$$

Point (1, 2) : role="math" localid="1646653269220" $$\begin{gathered} f\left(1\right)=2 \\ a·1^2+b·1+c=2 \\ a+b+5=2 \\ a+b=-3 \end{gathered}$$

Point (3, 2) : $$\begin{gathered} f\left(3\right)=2 \\ a·3^2+b·3+c=2 \\ 9a+3b+5=2 \\ 9a+3b=-3 \end{gathered}$$

Solve the equation to get the value of a and b from step 2.

$$\begin{gathered} a+b=-3 \\ 9a+3b=-3 \\ a+b=9a+3b \\ 8a+2b=0 \\ b=-\frac{8}{2}a=-4a \end{gathered}$$

Now put

$$\begin{gathered} b=-4a\mathrm{into}a+b=-3 \\ a+(-4a)=-3 \\ -3a=-3 \\ a=1 \end{gathered}$$

role="math" localid="1646654029072" $$\begin{gathered} a+b=-3 \\ 1+b=-3 \\ b=-4 \end{gathered}$$

So the quadratic function is:

role="math" localid="1646654053485" $$\begin{gathered} f\left(x\right)=a{x}^2+bx+c \\ f\left(x\right)=1·x^2+(-4)x+5 \\ f\left(x\right)=x^2-4x+5 \end{gathered}$$

The quadratic equation is$f\left(x\right)=x^2-4x+5$