Question

Form a quadratic function f(x) for which f(1)=0, f(−1)=−4, and f(2)=5.

Answer 1

Answer:

$f(x)=x^2+2x-3$

Step-by-step explanation:

$\textsf{General form of a quadratic function}:f(x)=ax^2+bx+c$

Equation 1

$\begin{aligned}f(1) &=0\\ \implies a(1)^2+b(1)+c &=0\\ a+b+c &=0 \end{aligned}$

Equation 2

$\begin{aligned}f(-1) &=-4\\ \implies a(-1)^2+b(-1)+c &=-4\\ a-b+c &=-4 \end{aligned}$

Equation 3

$\begin{aligned}f(2) &=5\\ \implies a(2)^2+b(2)+c &=5\\ 4a+2b+c &=5 \end{aligned}$

Add Equation 1 and Equation 2:

$\begin{array}{r l}a+b+c & =0 \\+\quad a-b+c & =-4 \\\cline{1-2} 2a+2c & =-4 \end{array}$

$\implies a+c=-2$

Substitute $a+c=-2$ into Equation 1 and solve for b:

$\begin{aligned}\implies a+b+c &=0\\ b-2 &=0\\ b &=2\end{aligned}$

Substitute $a=-2-c$  and  $b=2$  into Equation 3 and solve for c:

$\begin{aligned} \implies 4a+2b+c&=5\\ 4(-2-c)+2(2)+c &=5\\-8-4c+4+c &=5\\ -3c &=9\\ c &=-3\end{aligned}$

Substitute found value of c into $a=-2-c$ and solve for a:

$\implies a=-2-(-3)=1$

Therefore, a = 1, b = 2 and c = -3

Substitute the found values into the general form of a quadratic function to form the final equation:

$f(x)=x^2+2x-3$