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Section 3 Quadratic Functions and their modeling

In this section, we address the following course learning goals.

4Q

Be able to solve a quadratic equation.

5Q

Be able to determine the vertex and the equation of a quadratic function given its graph or a table of values.

6Q

Be able to model a situation with appropriate quadratic functions and interpret the solution including interpreting the vertex in context.

13F

Be able to determine inputs and outputs of a function from its graph and/or a table of values.

14F

Be able to determine the domain and range of function given as an equation or a graph.

18F

Be able to compute the average rate of change of a given function on a given interval.

20F

Be able to solve inequalities and interpret the solution.

21F

Be able to identify the intervals on which a given function is increasing or decreasing.

The function that gives the area of the rectangular region in terms of one side of the region is called a quadratic function. In general, a quadratic function is a function of the form

\begin{equation*} f(x) = ax^2+bx+c \end{equation*}

where \(a \neq 0\text{.}\) We call this the general form of a quadratic function. Now try to rewrite the equation that you used in Task 3.1 in the form of \(f(x)=ax^2+bx+c\text{.}\) What are the values of \(a, b\) and \(c\text{?}\)

If we want to find the function equation of a quadratic equation when given its graph, we have to pick three different points to plug into the formula \(f(x)=ax^2+bx+c\) because we have to figure out the values of three unknowns \(a,b\) and \(c\text{.}\) The computations will be lengthy because it involves solving a system of linear equations of three variables (it can be done pretty easily if you know how to use a computer package). We will not be doing this in this course. Instead, we will manipulate the function equation \(f(x)=ax^2+bx+c\) into a much desired form called the vertex form because a great deal of qualitative information about the function and its graph can simply be read off from this form. We will included the derivation of the algebraic manipulation in the following proof. Feel free to read the proof but it is not required.

Before we start doing the algebra, let us recall that

\begin{equation*} (X+Y)^2=X^2+2XY+Y^2. \end{equation*}

We will need this formula from (3.4) to (3.5). Now we manipulate the quadratic function from the general form into the vertex form.

\begin{align} f(x) \amp = ax^2+bx+c\tag{3.1}\\ f(x) \amp = ax^2+bx+\frac{b^2}{4a} - \frac{b^2}{4a}+c\tag{3.2}\\ f(x) \amp = a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}) - \frac{b^2}{4a}+c\tag{3.3}\\ f(x) \amp = a(x^2+2\frac{b}{2a}x + (\frac{b}{2a})^2) - \frac{b^2}{4a}+c\label{quadratic-step1}\tag{3.4}\\ f(x) \amp = a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c\label{quadratic-step2}\tag{3.5}\\ f(x) \amp = a(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a}\label{quadratic-goodform}\tag{3.6} \end{align}

By renaming \(h = -\frac{b}{2a}, k = -\frac{b^2-4ac}{4a},\) we have \(f(x)=a(x-h)^2+k\)

Take Equation (3.6), if we want to solve the quadratic equation

\begin{equation*} ax^2+bx+c = 0\text{,} \end{equation*}

it is enough to solve the equation

\begin{equation*} a(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a} = 0 \end{equation*}

This is easy to solve because we have

\begin{align} a(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a} \amp = 0\tag{3.7}\\ a(x+\frac{b}{2a})^2 \amp = \frac{b^2-4ac}{4a}\tag{3.8}\\ (x+\frac{b}{2a})^2 \amp = \frac{b^2-4ac}{4a^2}\tag{3.9}\\ x+\frac{b}{2a} \amp = \pm \sqrt{\frac{b^2-4ac}{4a^2}}\tag{3.10}\\ x \amp = -\frac{b}{2a} \pm \sqrt{\frac{b^2-4ac}{4a^2}} \tag{3.11}\\ x \amp = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\label{quadratic-formula}\tag{3.12} \end{align}

Equation (3.12) is the well-known quadratic formula that you have seen before.

The vertex form of a quadratic equation is of the form

\begin{equation*} f(x)=a(x-h)^2+k\text{.} \end{equation*}

The reason this is called the vertex form is because we can read off the vertex of the graph, which is \((h,k)\text{.}\) The vertex of a quadratic function is the lowest point (if \(a \gt 0\)) or the highest point (if \(a \lt 0\)). Using the vertex form of a quadratic equation, we can find the function equation of a quadratic equation by first identifying the vertex (so we will know the values of \(h\) and \(k\)), and then pick a point on the graph of the quadratic equation other than the vertex to find the value of \(a\text{.}\)

Definition 3.4.
The domain of a function is the set of allowable inputs to the function. The range of a function is the set of all possible outputs from the function.
Definition 3.5.
A function \(y=f(x)\) is said to be increasing if the graph of \(f(x)\) is going uphill from left to right. If the graph of \(f(x)\) is going downhill from left to right, then it is called decreasing.
Definition 3.8.
The average rate of change of a function \(f(x)\) defined on an interval \(a \le x \le b\) is
\begin{equation*} \frac{f(b)-f(a)}{b-a}. \end{equation*}

Note that the average rate of change computes the slope of the line between the two points \((a,f(a))\) and \((b,f(b))\text{.}\)