Intervals and Secants

Suppose we have a qudratic function $Q(x) = a (x - h)^2 + k$, with $a > 0$.

Then its graph is a parabola.

Let’s examine the interval $[2, 5]$.

$\blacktriangleright$ Algebraically, the rate of change of $Q(x)$ over the interval $[2,5]$ is given by

$$\frac {Q(5) - Q(2)}{5 - 2}$$

$\blacktriangleright$ Geometrically, this is the slope of the secant line running through the points $(2, Q(2))$ and $(5, Q(5))$.

Intervals and secants are algebraic and geometric partners. Secants give a picture of rates of change over an interval.

What if we push the secant over a little bit until it becomes a tangent line?

Now it is a picture of the rate of change over the interval $\left [ \tfrac {7}{2}, \tfrac {7}{2} \right ]$.

How should we interpret this?

What is the rate of change AT a point?

You cannot calculate the rate of change over an interval with $0$ length. Therefore, we will invent an interpretation for this rate of change at a point, i.e. an instantaneous rate of change.

Tangent Lines

Suppose $f$ is a function. The graph of $f$ is the collection of points whose coordinates are pairs in $f$. That is, their coordinates look like $(d, f(d))$.

Suppose $a$ and $b$ are two distinct domain numbers of $f$. Then the line through $(a, f(a))$ and $(b, f(b))$ is called a secant line. The slope of this secant line equals the rate of change of $f$ over the interval $[a, b]$.

Tangent lines are degenerate secants. Secant lines need two points on the graph of a function. A tangent line is a secant line where the two points are the same point. Tangent lines are secant lines created from an interval of length $0$. But tangent lines are still lines. They still have a slope.

If the slope of a secant line corresponds to the function’s rate of change over an interval, then the slope of a tangent line corresponds to the function’s rate of change at a single number.

We’ll call this the instantaneous rate of change at the domain number corresponding to the tangent point.

Quadratics

We’ll begin our investigation into instantaneous rate of change with quadratic functions, whose graphs are parabolas (which never have vertical tangent lines).

We need a method of obtaining the slope of tangent lines to parabolas.

$\blacktriangleright$ Basic Quadratic

We’ll begin with our basic quadratic function: $Q(x) = x^2$ and its graph $y = x^2$.

We are investigating the graph of $y = x^2$, which is a parabola whose vertex is $(0, 0)$.

Let’s select a domain number of $Q$, call it $x_0$. This corresponds to the point $(x_0, y_0) = (x_0, (x_0)^2)$ on the graph of $Q$.

Let’s add a picture of the tangent line at $(x_0, y_0) = (x_0, (x_0)^2)$ to our graph of $Q$.

The tangent line is the graph of a linear function. Let’s call this linear function $T$.

$T$ is a linear function. Therefore, the formula for $T$, would look like $T(x) = m(x - x_0) + y_0$, for some $m$. According to our definition, $m$ is the instantaneous rate of change of $Q$ at $x_0$. How do we determine its exact value?

$\blacktriangleright$ We would like a way to obtain the exact value of $m$.

To get the exact value of $m$, we are going to create a new quadratic function.

At any point $(x_0, y_0)$ on the graph of $y = x^2$, the slope of the tangent line is $2 \, x_0$. The slope of the tangent line is always twice the $x$-coordinate of the tangent point.

$\blacktriangleright$ Stretching

We can vertically stretch our basic quadratic parabola by multiplying all of the $y$-coordinates by a constant, $a \ne 0$.

The new parabola has an equation of the form $y = a \, x^2$.

These are graphs of quadratic functions of the form $Q(x) = a \, x^2$.

Let’s go through the same process.

At any point $(x_0, y_0)$ on the graph of $y = a \, x^2$, the slope of the tangent line is $2 \, a \, x_0$. The slope of the tangent line is always $2a$ times the $x$-coordinate of the tangent point.

$\blacktriangleright$ Shifting

Beginning with $Q(x) = a \, x^2$ and its parabola $y = a \, x^2$, we can shift horizontally and vertically.

If we keep the shape of the parabola, $y = a \, x^2$, intact and shift its position, then the new parabola has an equation of the form $y = a \, (x - h)^2 + k$.

These are graphs of quadratic functions of the form $Q_{new}(x) = a \, (x - h)^2 + k$.

If we pick any point, $(x_0, y_0)$, on this parabola, then the slope of the tangent line is the same as the slope of the tangent line for $y = a \, x^2$ at the point $(x_0 - h, y_0 - k)$.

From above, we know this slope is $2 \, a \, (x_0 - h)$.

Note: If we look at the vertex $(x_0, y_0) = (h, k)$, then $2 a (x_0 - h) = 2 a (h - h) = 0$ and the slope of the tangent line is $0$, just like we had reasoned before.

iRoC

Let $Q(x) = a (x - h)^2 + k$ be any quadratic function.

Every point, $(x_0, y_0)$, on the graph of $y = Q(x)$ has a tangent line.

Each tangent line has a slope, which we are calling the instantaneous rate of change of $Q$ at $x_0$.

$\blacktriangleright$ We can create a new function from this.

We now have a function, with a formula, for measuring the instantaneous rate of change of any quadratic function at any domain number.

$\blacktriangleright$ The instantaneous rate of change of the quadratic function, $Q(x) = a \, (x - h)^2 + k$, is the linear function, $iRoC_Q(x) = 2 \, a \, (x-h)$.

We already have a function, with a formula, for measuring the instantaneous rate of change of any linear function at any domain number.

$\blacktriangleright$ The instantaneous rate of change of the linear function, $L(x) = m \, x + b$, is the constant function, $iRoC_L(x) = m$.

$\blacktriangleright$ The instantaneous rate of change of the constant function, $C(x) = c$, is the zero function, $iRoC_C(x) = 0$.

We will use both names. This will keep the idea fresh in our heads and also give us experience with Calculus notation.