Find the Extreme Value
The extreme value of a quadratic function \(f(x)\) is either the minimum or maximum value of the quadratic function \(f(x)\). The extreme value is given as
\[f\left(-\dfrac{b}{2a}\right)\nonumber\]
The minimum value is located at the lowest point of an upwards parabola and the maximum value is located at the highest point of a downwards parabola.
Notice in the definition that the extreme value is the \(y\)-coordinate of the vertex. Recall, the vertex is
\[\left(-\dfrac{b}{2a},f\left(-\dfrac{b}{2a}\right)\right)\nonumber\]
and we can see the \(y\)-coordinate is just the extreme value. Hence, when we use the words vertex, minimum or maximum value, or extreme value, they are all associated with the vertex of a parabola.
Find the vertex and the extreme value of the function \(f(x) = x^2 − 4x + 5\).
Solution
Notice \(a = 1\), which means \(a > 0\). Hence, \(f(x)\) is an upwards parabola and, from the definition, we expect \(f(x)\) to have a minimum value. Let’s use the formula to find the vertex, where \(a = 1\) and \(b = −4\).
\[\begin{array}{rl}x=-\dfrac{b}{2a}&\text{Plug-n-chug} \\ x=-\dfrac{\color{blue}{-4}}{\color{blue}{2(1)}}&\text{Simplify} \\ x=2&\text{The }x\text{-coordinate of the vertex}\end{array}\nonumber\]
Next, we find the \(y\)-coordinate of the vertex by obtaining \(f(2)\).
\[\begin{array}{rl}f(x)=x^2-4x+5&\text{Plug-n-chug} \\ f(\color{blue}{2}\color{black}{)}=(\color{blue}{2}\color{black}{)}^2-4(\color{blue}{2}\color{black}{)}+5&\text{Evaluate} \\ f(2)=1&\text{The }y\text{-coordinate of the vertex}\end{array}\nonumber\]
Hence, the vertex is \((2, 1)\). Next, we find the extreme value of \(f(x)\). From the vertex calculation, we see \(f\left(-\dfrac{b}{2a}\right)=f(2)=1\). Thus, the extreme value is \(1\), the \(y\)-coordinate of the vertex.
Find the vertex and the extreme value of the function \(q(n) = −3n^2 − 5n + 3\).
Solution
Notice \(a = −3\), which means \(a < 0\). Hence, \(q(n)\) is an downwards parabola and, from the definition, we expect \(q(n)\) to have a maximum value. Let’s use the formula to find the vertex, where \(a = −3\) and \(b = −5\).
\[\begin{array}{rl}n=-\dfrac{b}{2a}&\text{Plug-n-chug} \\ n=-\dfrac{\color{blue}{-5}}{\color{blue}{2(-3)}}&\color{black}{\text{Simplify}} \\ n=-\dfrac{5}{6}&\text{The }x\text{-coordinate of the vertex}\end{array}\nonumber\]
Next, we find the \(y\)-coordinate of the vertex by obtaining \(q\left(-\dfrac{5}{6}\right)\).
\[\begin{array}{rl} q(n)=-3n^2-5n+3&\text{Plug-n-chug} \\ q\left(\color{blue}{-\dfrac{5}{6}}\color{black}{}\right)=-3\left(\color{blue}{-\dfrac{5}{6}}\color{black}{}\right)^2-5\left(\color{blue}{-\dfrac{5}{6}}\color{black}{}\right) +3&\text{Evaluate} \\ q\left(-\dfrac{5}{6}\right)=\dfrac{61}{12}&\text{The }y\text{-coordinate of the vertex}\end{array}\nonumber\]
Hence, the vertex is \(\left(-\dfrac{5}{6},\dfrac{61}{12}\right)\). Next, we find the extreme value of \(q(n)\). From the vertex calculation, we see \(q\left(-\dfrac{b}{2a}\right)=q\left(-\dfrac{5}{6}\right)=\dfrac{61}{12}\). Thus, the extreme value is \(\dfrac{61}{12}\), the \(y\)-coordinate of the vertex.
It may seem a little redundant to find the vertex and the extreme value, but, remember, the goal is to apply this concept to real-world applications with quadratic functions. Let’s take a look at a few applications where we find the extreme value in context of a real-world model.