Question

How do you find a quadratic function $f\left( x \right)=a{{x}^{2}}+bx+c$ given a maximum value of 9 and the roots of the equation are – 6 and 0.

Answer 1

Hint: Now note that the roots of the equation are – 6 and 0. We know that the sum of roots is given by $\dfrac{-b}{a}$ hence we will get the relation between a and b. Now we know that the product of the roots is given by $\dfrac{c}{a}$ hence we will get the value of c in the equation. Now know that the maximum value of quadratic equation is given by $c-\dfrac{{{b}^{2}}}{4a}$ . hence we will substitute the value of b and c and find the value of a. Now we will substitute the value of a and c to find the value of b and Hence we can now write the whole quadratic equation.

Complete step by step solution:
Now we are given that -6 and 0 are the roots of quadratic equation $a{{x}^{2}}+bx+c$
Now we know that the sum of roots of quadratic equation is given by $\dfrac{-b}{a}$
Hence we have $-6+0=\dfrac{-b}{a}$
Which means $b=6a...............\left( 1 \right)$
Now we also know that the product of the roots is given by $\dfrac{c}{a}$ .
Similarly we have $-6\times 0=\dfrac{c}{a}$ Hence we get c = 0.
Now we know that the maximum value of a quadratic function is $c-\dfrac{{{b}^{2}}}{4a}$
We are given that the maximum value is 9. Hence we get,
$\begin{align}
  & \Rightarrow c-\dfrac{{{b}^{2}}}{4a}=9 \\
 & \Rightarrow 0-\dfrac{{{\left( 6a \right)}^{2}}}{4a}=9 \\
 & \Rightarrow -\dfrac{36{{a}^{2}}}{4a}=9 \\
 & \Rightarrow -9a=9 \\
 & \Rightarrow a=-1 \\
\end{align}$
Hence we get the value of a is – 1
Now from equation (1) we have $b=6\left( -1 \right)=-6$
Now we have c = 0 hence we get the quadratic $a{{x}^{2}}+bx+c$ as $-{{x}^{2}}-6x$

Note:
Now note that we can avoid using sum of the roots and product of the roots to solve the problem. Since -6 and 0 are the roots of the quadratic and we know that the roots satisfy the equation $a{{x}^{2}}+bx+c=0$ we will substitute x = 0 and x = -6 to find the required condition. Now again we will use the condition of maxima to find the value of a and hence write the whole quadratic equation. Hence we get the required quadratic equation.