Quadratic Functions, Graphing, and Optimization

Introduction to Functions and Graphs

There is a distinct difference between traditional equation notation and function notation, though they are often used to represent the same relationships. Equation notation typically takes the form of $y = ax^2 + bx + c$, whereas function notation is expressed as $f(x) = ax^2 + bx + c$. Function notation offers two primary benefits. First, it is clearer in text-based descriptions; for instance, stating that the height of a ball depends on time can be succinctly noted as $h(t)$. Second, it provides clearer labeling for graphs. When graphing a function, creating a table of values is a helpful starting point. For example, to graph $f(x) = x^2$, one should create a table with at least five coordinate pairs, such as $x = -2, -1, 0, 1, 2$ and their corresponding output values of $y = 4, 1, 0, 1, 4$.

When constructing graphs, it is important to note that while choosing which $x$ values to pick can be challenging at first, this skill becomes intuitive over time. Axes must always be properly scaled and labeled; if a scale is not explicitly labeled, it is assumed to be a ratio of $1:1$. The shape resulting from a quadratic function is called a parabola, and the turning point of this shape is known as the vertex. From the vertex, the base pattern of a parabola exhibits specific movements: moving one unit left or right from the vertex equates to moving one unit up ($1^2$), and moving two units left or right equates to moving four units up ($2^2$). This pattern is essential to remember for later work with transformations.

Transforming Parabolas and the Role of Variables

Transformations of the parent function $f(x) = x^2$ can be analyzed by varying the parameters $a$, $h$, and $k$ in the vertex form: $f(x) = a(x - h)^2 + k$. In the function $a(x) = x^2 + 1$, the vertex shifts upward. In $b(x) = (x + 1)^2$, the graph shifts horizontally. In $c(x) = (x + 1)^2 + 1$, both vertical and horizontal shifts occur. When the coefficient $a$ is modified, such as in $d(x) = 2x^2$ or $e(x) = \frac{1}{2}x^2$, the width and direction of the parabola change. Specifically, the variable $a$ determines the direction (upward or downward) and the vertical stretch (thin or wide). The variable $h$ determines the horizontal translation (shifting left or right), and the variable $k$ determines the vertical translation (shifting up or down).

Completing the Square and Vertex Form

To graph a quadratic function effectively, it is often useful to convert the general form into the vertex form, $y = a(x - h)^2 + k$, by using the process of completing the square. In this form, the vertex is located exactly at the coordinates $(h, k)$. For the example $y = 2x^2 - 4x - 7$, the transformation steps are as follows. First, factor out the $a$ coefficient from the first two terms: $y = 2(x^2 - 2x) - 7$. Second, add and subtract the square of half the coefficient of the $x$ term inside the parenthesis: $y = 2(x^2 - 2x + 1 - 1) - 7$. Third, rewrite the right-hand side to isolate the perfect square trinomial using the commutative property: $y = 2(x^2 - 2x + 1) - 7 - 2$. Note that the $-2$ comes from multiplying the subtracted $1$ by the lead coefficient $2$. Fourth, factor and simplify: $y = 2(x - 1)^2 - 9$. This reveals the vertex is at $(1, -9)$ and the parabola points upward.

To complete a graph, one must also identify the intercepts. The $y$-intercept occurs when $x = 0$; for this function, yielding the point $(0, -7)$. The $x$-intercepts occur when $y = 0$. By setting the vertex form to zero, we find $0 = 2(x - 1)^2 - 9$, which leads to $9 = 2(x - 1)^2$, then $\frac{9}{2} = (x - 1)^2$. Taking the square root gives $\frac{\text{3}}{\text{sqrt(2)}} = x - 1$, resulting in $x = 1 \pm \frac{3\text{sqrt}(2)}{2}$, or approximately $x ≈ 3.12$ and $x ≈ -1.12$.

Intercept Form and Symmetry

There is a third form to consider when graphing quadratics, known as the Intercept Form or Root Form: $y = a(x - r_1)(x - r_2)$. Parabolas are symmetrical, and this symmetry is defined by the Axis of Symmetry, which is the vertical line over which the parabola is mirrored. The equation for this line is $x = \frac{r_1 + r_2}{2}$, which is the midpoint of the roots or intercepts. This line always passes through the vertex. Therefore, if the axis of symmetry is known, the $x$-coordinate of the vertex is also known.

For the example $f(x) = 2x^2 - 28x + 96$, we can factor out the constant to get $f(x) = 2(x^2 - 14x + 48)$, which factors further into $y = 2(x - 6)(x - 8)$. The roots are located at $x = 6$ and $x = 8$. The line of symmetry is at the midpoint, $x = 7$. To find the $y$-coordinate of the vertex, substitute the symmetry value back into the equation: $y = 2(7 - 6)(7 - 8) = 2(1)(-1) = -2$. Thus, the vertex is identified as $(7, -2)$. The discriminant ($D$) of a quadratic equation can also tell us the number of solutions graphically. If $D > 0$, there are two real solutions (two intercepts). If $D = 0$, there is one real solution (the vertex sits on the axis). If $D < 0$, there are no real solutions (the graph does not touch the $x$-axis).

Vertex and Optimization

The vertex of a parabola always represents the maximum or minimum value of the quadratic function. If a problem asks to find the maximum or minimum of a variable, such as cost, height, or profit, it is a direct clue to look for the vertex. Consider a practical application involving a rectangular field where three sides are to be surrounded by $400 \text{m}$ of fencing, and the fourth side is enclosed by an adjacent barn. To find the dimensions that give the maximum area, we define the perimeter constraint as $2l + w = 400$ and the area as $A = lw$. By substituting the perimeter into the area equation, we get $A = l(400 - 2l) = -2l^2 + 400l$.

Using completing the square on this area function, we factor out $-2$ to get $-2(l^2 - 200l)$, then add and subtract $10,000$ inside the parenthesis. This results in $A = -2(l - 100)^2 + 20,000$. The vertex is at $(100, 20,000)$, meaning the maximum area is $20,000 \text{m}^2$. Alternatively, using the intercept form, the roots for $A = -2l(l - 200)$ are $l = 0$ and $l = 200$, placing the vertex at the midpoint $l = 100$. The optimal dimensions are therefore $100 \text{m}$ by $200 \text{m}$. When solving optimization problems, always check whether the question asks for the dimensions, the area, or both.