Question

How do you determine the range of a quadratic function?

Answer 1

Hint: In this question, we have to find the range of a quadratic function. As we know, a quadratic function is a function which has one variable and has a power of 2. Also, range consists of the values, where the output of a function lies. Thus, in this problem, we let the general form of the quadratic equation $f(x)=a{{x}^{2}}+bx+c$ , and then we will find the first derivative of the given equation. After that, we will put the derivative equal to 0, to get the value of x. Then, we will find the second derivative of the function. In the last, we will substitute the value of x in the quadratic equation, to get the required range of the solution.

Complete step-by-step solution:
According to the question, we have to find the range of a quadratic polynomial. As we know, the range is the interval where the value of output lies.
Thus, we will let the general form of the quadratic equation be $f(x)=a{{x}^{2}}+bx+c$, where a, b, and c are the real numbers and x is the variable. ---- (1)
Now, we will find the first derivative of equation (1) with respect to x, we get
$\Rightarrow f'(x)=2ax+b$ --------- (2)
Now, we will put the value of first derivative equal to 0 that is let f’(x) = 0 to get the value of x, we get
$\Rightarrow 2ax+b=0$
Now, subtract b on both side in the above equation, we get
$\Rightarrow 2ax+b-b=0-b$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow 2ax=-b$
Now, we will divide 2a on both sides in the above equation, we get
$\Rightarrow \dfrac{2a}{2a}x=\dfrac{-b}{2a}$
On further simplification, we get
$\Rightarrow x=\dfrac{-b}{2a}$ -------- (3)
Now, we will again find the derivative of equation (2), we get
$\Rightarrow f''(x)=2a$
Since, 2a is a positive number; it implies the value of the function at $x=\dfrac{-b}{2a}$ is minimum.
So, we will substitute the value of x from equation (3) in equation (1), we get
$\Rightarrow f\left( \dfrac{-b}{2a} \right)=a{{\left( \dfrac{-b}{2a} \right)}^{2}}+b\left( \dfrac{-b}{2a} \right)+c$
On solving the brackets of the above equation, we get
$\Rightarrow f\left( \dfrac{-b}{2a} \right)=\dfrac{a{{b}^{2}}}{4{{a}^{2}}}-\dfrac{{{b}^{2}}}{2a}+c$
On further simplifying the above equation, we get
$\Rightarrow f\left( \dfrac{-b}{2a} \right)=\dfrac{{{b}^{2}}}{4a}-\dfrac{{{b}^{2}}}{2a}+c$
On taking the LCM of the denominator on the right-hand side in the above equation, we get
$\Rightarrow f\left( \dfrac{-b}{2a} \right)=\dfrac{{{b}^{2}}-2{{b}^{2}}+4ac}{4a}$
On further solving, we get
$\Rightarrow f\left( \dfrac{-b}{2a} \right)=\dfrac{-{{b}^{2}}+4ac}{4a}$
Therefore, the minimum limit of the range of the function is $f\left( \dfrac{-b}{a} \right)$ and maximum limit is $\infty $
Thus, the range of a quadratic equation is $\left( f\left( \dfrac{-b}{a} \right),\infty \right)$ , that is $\left( \dfrac{-{{b}^{2}}+4ac}{4a},\infty \right)$.

Note: While solving this problem, do mention the formula you are using to avoid confusion and mathematical error. Always remember that when the value of the second derivative is negative, it means the value of the function is maximum, otherwise it is minimum.