Question

Give an example of a quadratic function that has a maximum value. How do you know that it has maximum value?

Answer 1

By inspection, we get to know that a function whose graph opens downward has a maximum value. Use the standard form of quadratic expression $a{{x}^{2}}+bx+c$ where $a\ne 0$ to find the maximum value of a quadratic function. When $a<0,$ the quadratic expression has a maximum value. We will use the discriminant test to show this.

Complete step by step solution:
An example of a quadratic function that has a maximum value is given by the equation, $y=-{{x}^{2}}.$
This quadratic function is the equation of a parabola whose graph opens downward.
That is, in this quadratic equation $a=-1<0.$ Hence, this quadratic function, which is a parabola, has a maximum value.
Let us prove that a quadratic expression of the form $a{{x}^{2}}+bx+c$ with $a<0$ has a maximum value.
Consider the standard quadratic expression $a{{x}^{2}}+bx+c$ with $a\ne 0$ for all real values of $x.$
Let us suppose that $y=a{{x}^{2}}+bx+c.$
Then we will get, $a{{x}^{2}}+bx+c-y=0$ by transposing $y$ from the left-hand side to the right-hand side.
We use the discriminant test to show that the quadratic expression given here has a maximum value.
Since $x$ is a real number, the discriminant of the quadratic equation $a{{x}^{2}}+bx+c-y=0$ is greater than or equal to zero.
The discriminant of the above expression is given by ${{b}^{2}}-4a\left( c-y \right).$
So, we get,
$\Rightarrow {{b}^{2}}-4a\left( c-y \right)\ge 0.$
Let us open the bracket in the above inequality, we get
$\Rightarrow {{b}^{2}}-4ac+4ay\ge 0.$
Now, we are going to transpose the terms without $y$ from the left-hand side to the right-hand side.
We get,
$\Rightarrow 4ay\ge -\left( {{b}^{2}}-4ac \right).$
That is,
$\Rightarrow 4ay\ge 4ac-{{b}^{2}}.$
Suppose that $a<0.$
Transpose $4a$ from the left-hand side to the right-hand side, we get
$\Rightarrow y\le \dfrac{4ac-{{b}^{2}}}{4a}.$ Since $a<0,$ the inequality changes.
The above inequality proves that $y$ has a maximum value $\dfrac{4ac-{{b}^{2}}}{4a}.$

Hence, the quadratic function $y=-{{x}^{2}}$ has a maximum value, since $a=-1 <0.$

Note: If we substitute the value $y=\dfrac{4ac-{{b}^{2}}}{4a}$ in the equation $a{{x}^{2}}+bx+c-y=0$ we get the $x$-coordinate of the point at which this expression has a maximum value.
So, $a{{x}^{2}}+bx+c-\dfrac{4ac-{{b}^{2}}}{4a}=4{{a}^{2}}{{x}^{2}}+4abx+4ac-4ac+{{b}^{2}}=0$
$\Rightarrow 4{{a}^{2}}{{x}^{2}}+4abx+{{b}^{2}}={{\left( 2ax+b \right)}^{2}}=0$
$\Rightarrow 2ax+b=0$
$\Rightarrow x=\dfrac{-b}{2a}.$
Therefore, the expression $y=a{{x}^{2}}+bx+c$ has its maximum value at $x=\dfrac{-b}{2a}.$
Thus, the $x$-coordinate of the point at which the function $y=-{{x}^{2}}$ has its maximum value is $x=0,$ since $b=0.$
The point at which this function has maximum value is $\left( 0,0 \right),$ since $b=c=0.$