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Section 1.1 Intro to Quadratic Functions

Previously, most of the relationships between variables were linear.

Subsection 1.1.1 Motivation: Galileo's Study of Freefall Motion

Consider the motion of an object which is dropped, and allowed to fall freely, only under the influence of gravity.

In the 16th century, Galileo Galilei (1564-1642) studied how objects falls to the Earth, and tried to quantify how they fall over time. At the time, it was difficult to measure the motion of a falling rock, because they accelerate and fall very quickly. Then, capturing its motion would require a very accurate clock, which were not developed at the time. Instead, Galileo studied the motion of a ball rolling down a ramp (an inclined plane) with only a small slope, so that the ball accelerates slowly over time. He used a ramp which was smooth, and a ball which was hard, smooth, and round, in order to avoid friction. This created an idealized situation to try and capture the essence of the motion.

To time the ball's decent, he used a water clock, which works by opening a valve to start the clock, water flowing at a constant rate into a container, and closing the value to stop the clock. Then, weighing the amount of water allows you to measure the amount of time elapsed. He conducted a variety of different experiments, varying the slope of the ramp, varying the distance travelled by the ball, etc.

Galileo found that, “the distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity”. This means that if the ball rolls \(d\) units in the first unit of time, then it rolls an additional \(3d\) units in the second unit of time, then \(5d\) units, then \(7d\text{,}\) and so on.

Then, if the ball travels 1 unit of distance in the 1st unit of time, then total distance travelled by the ball is after 2 units of time is \(1 + 3 = 4\) units of distance. After 3 units of time, \(1 + 3 + 5 = 9\) units. In general,

\begin{equation*} \begin{array}{c|c} \text{time} \amp \text{distance} \\ \hline 1 \amp 1 \\ 2 \amp 1 + 3 = 4 \\ 3 \amp 1 + 3 + 5 = 9 \\ 4 \amp 1 + 3 + 5 + 7 = 16 \end{array} \end{equation*}

Notice that the distances are the squares of the time. In modern mathematics, we say that the distance travelled is (directly) proportional to the square of time. Let \(d\) represent the distance travelled, \(t\) represent the time elapsed. Then,

\begin{equation*} d = k t^2 \end{equation*}

for some proportionality constant \(k\text{,}\) which depends on the units of \(d\) and \(t\text{.}\) If \(d\) is measured in meters, and \(t\) is measured in seconds, then the equation turns out to be,

\begin{equation*} d = 4.9 t^2 \end{equation*}

That is, \(t\) seconds after dropping an object, the object will have travelled \(4.9 t^2\) meters.

This Desmos applet shows the graph of motion, where the horizontal axis represents time, and the vertical axis is the distance travelled. The graph of this relationship forms a curve called a parabola.

Next, consider the height \(h\) of an object in freefall motion. If the object is initially dropped from a height of \(h_0\text{,}\) then at \(t = 0\text{,}\) \(h = h_0\text{.}\) As the object falls, it travells a distance of \(d = 4.9 t^2\) downwards, which decreases its height. Thus, the height \(h\) of the dropped object is given by,

\begin{equation*} h = h_0 - 4.9 t^2 \end{equation*}

This produces a parabola that opens downward, as opposed to the previous example which opened upward. Consider the applet, and consider how the graph changes when the initial height \(h_0\).

Further, if the object is initially thrown upward with initial velocity \(v_0\text{,}\) then the height of the object turns out to be,

\begin{equation*} h = h_0 + v_0 t - 4.9 t^2 \end{equation*}

That is, a linear term \(v_0 t\) is added to the equation. Here, \(h\) is a quadratic function of \(t\text{.}\) Again, consider the applet, and see how the graph of motion changes by varying the inital velocity \(v_0\) and initial height \(h_0\).

Subsection 1.1.2 Quadratic Functions

The operation of squaring a number, that is, going from \(x\) to \(x^2\text{,}\) is a function. The function \(f(x) = x^2\) is called the squaring function.

Recall that the area \(A\) of a square with side length \(x\) is given by \(A = x^2\text{,}\) a squaring relationship.

Definition 1.1.2.

A quadratic function is a function of the form,

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

where \(a, b, c \in \mathbb{R}\) are real numbers, and \(a \neq 0\text{.}\)

The term “quadratic” comes from the Latin “quadratus” meaning “square”, referring to the fact that the independent variable \(x\) is squared.

This is called the general form of a quadratic function. The domain of a quadratic function is all real numbers, \(x \in \mathbb{R}\) or \((-\infty,\infty)\text{.}\)

Subsection 1.1.3 Graph of a Quadratic Function

The graph of a quadratic function is a curve called a parabola. It has the shape of a “bowl”, or a U-shape. A parabola has particular characteristics. Consider a quadratic function of the form \(y = ax^2 + bx + c\text{.}\)

  • Opening up/down. If the coefficient \(a\) on the \(x^2\) term is positive (\(a > 0\)), then the parabola “opens up”. If this coefficient is negative (\(a \lt 0\)), the parabola “opens down”. This number \(a\) is called the leading coefficient, because it is the coefficient of the highest power of \(x\text{.}\)

  • Vertex. The vertex of a parabola is its maximum point or minimum point. If the parabola opens up, the vertex is a minimum point, and if it opens down, the vertex is a maximum point.

  • Axis of symmetry. Parabolas are symmetric, in that they are made of up of two “halves” which are mirror images of each other. In particular, they are symmetric about the vertical line which passes through the vertex of the parabola, called the axis of symmetry.

The range of a quadratic function. If the vertex \((h,k)\) is a minimum point, then the range is \([k, \infty)\text{.}\) If the vertex is a maximum point, then the range is \((-\infty, k]\text{.}\)

Subsection 1.1.4 Intercepts of a Quadratic Function

For a quadratic function in general form, the \(y\)-intercept is given by,

\begin{equation*} f(0) = a(0)^2 + b(0) + c = c \end{equation*}

so it is the \(c\text{,}\) or the point \((0,c)\text{,}\) i.e. the constant term of the polynomial.

The \(x\)-intercept(s) can be determined by solving the equation \(f(x) = 0\text{,}\) i.e. they correspond to the solutions to the equation,

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

This is called a quadratic equation. Solving a quadratic equation by-hand is more complicated than solving linear equations, and various techniques will be developed in upcoming sections.

A solution to the equation are called a zero, or root. The terms “\(x\)-intercept”, “zero”, and “root”, are all somewhat synonyms. Sometimes, \(x\)-intercepts refers specifically to the points in the plane where the function passes through the \(x\)-axis whereas “zeros” or “roots” refer to the \(x\)-values of the function such that the function is 0.