Question

Find a quadratic function with the given zeros and passing through the given point.

Answer 1

The Solution.

The given zeros of the quadratic function are

$x=\frac{1}{7},x=0$

This implies that

$\begin{gathered} x=\frac{1}{7} \\ \text{cross multiplying, we get} \\ 7x=1 \\ 7x-1=0 \\ \text{ Similarly, x=0} \end{gathered}$

The required quadratic function can be obtained as below:

$\begin{gathered} (7x-1)x=0 \\ \text{Clearing the bracket, we get} \\ f(x)=7x^2-x=0 \end{gathered}$

So, the required quadratic function is

$\begin{gathered} f(x)=a(7x^2-x=0)\ldots\text{eqn}(1) \\ \text{where a is a constant, to be determined.} \end{gathered}$

To find the value of a, we shall apply the given initial values, that is,

$(4,-3)\Rightarrow x=4,f(x)=-3$

We get,

$-3=a\lbrack7(4^2)-4\rbrack$$\begin{gathered} -3=a(7\times16-4) \\ -3=a(112-4) \end{gathered}$$\begin{gathered} -3=a(108) \\ -3=108a \\ \text{Dividing both sides by 108, we get} \\ a=-\frac{3}{108} \\ \\ a=-\frac{1}{36} \end{gathered}$$\text{ Substituting -}\frac{1}{36}\text{ for a in eqn(1) above, we get}$$\begin{gathered} f(x)=-\frac{1}{36}(7x^2-x=0) \\ \\ f(x)=-\frac{7}{36}x^2+\frac{1}{36}x=0\text{ } \\ Or \\ \text{Multiplying through the equation by 36, we get} \\ 7x^2+x=0 \end{gathered}$

Hence, the correct quadratic function is

$\begin{gathered} f(x)=-\frac{7}{36}x^2+\frac{1}{36}x=0 \\ Or \\ f(x)=7x^2+x=0 \end{gathered}$