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M6.2 Slope of a Nonlinear Function

Figure M6.2 provides a graph of the function

A graph illustrates quadratic function. — Figure M6.2 Graph of Quadratic Function

Y = X^{2} - 4 X + 6

$Y = X^{2} - 4 X + 6$

Notice that this is not a straight line. To determine the slope of a curve at any point, we find the slope of a line tangent to the curve at this point. For example, if we wish to find the slope of the line when $X = 3,$ $X = 3,$ we find a line that is tangent at that point. To find the slope of a line, we use Equation M6-2, which requires us to have two points. We want the slope at $X = 3,$ $X = 3,$ so we let $X_{1} = 3$ $X_{1} = 3$ and we find

Y_{1} = {(3)}^{2} - 4 (3) + 6 = 3

$Y_{1} = {(3)}^{2} - 4 (3) + 6 = 3$

Thus, one point will be (3,3). We wish to find another point close to this, so we will choose a value of $X_{2}$ $X_{2}$ close to $X_{1} = 3.$ $X_{1} = 3.$ If we pick $X_{2} = 5$ $X_{2} = 5$ for the other point (an arbitrary choice used simply to illustrate this process), we find

Y_{2} = {(5)}^{2} - 4 (5) + 6 = 11

$Y_{2} = {(5)}^{2} - 4 (5) + 6 = 11$

This other point is (5,11).

The slope of the line between these two points is

b = \frac{Δ Y}{Δ X} = \frac{11 - 3}{5 - 3} = \frac{8}{2} = 4

$b = \frac{Δ Y}{Δ X} = \frac{11 - 3}{5 - 3} = \frac{8}{2} = 4$

To get a better estimate of the slope where $X_{1} = 3,$ $X_{1} = 3,$ we choose a value of $X_{2}$ $X_{2}$ even closer to $X_{1} = 3.$ $X_{1} = 3.$ Taking the point (3,3) with the point (4,6), we have

b = \frac{Δ Y}{Δ X} = \frac{6 - 3}{4 - 3} = \frac{3}{1} = 3

$b = \frac{Δ Y}{Δ X} = \frac{6 - 3}{4 - 3} = \frac{3}{1} = 3$

Neither of these slopes would be exactly the same as the slope of the line tangent to this curve at the point (3,3), as shown in Figure M6.3. However, we see that they each give us an estimate of the slope of a tangent line. If we keep selecting points closer and closer to the point where $X = 3,$ $X = 3,$ we find slopes that are closer to the value we are trying to find.

A graph illustrates tangent lines and other lines connecting the points. — Figure M6.3 Graph of Tangent Line and Other Lines Connecting Points

Figure M6.3 Full Alternative Text

To get a point very close to (3,3), we will use a value, $Δ X,$ $Δ X,$ and add this to $X = 3$ $X = 3$ to get the second point. As we let $Δ X$ $Δ X$ get smaller, we find a point $(X + Δ X, Y_{2})$ $(X + Δ X, Y_{2})$ closer to the original point (3,3). From the original equation,

Y = X^{2} - 4 X + 6

$Y = X^{2} - 4 X + 6$

we have

Y_{1} = 3^{2} - 4 (3) + 6 = 3

$Y_{1} = 3^{2} - 4 (3) + 6 = 3$

and

\begin{array}{l} Y_{2} & = & {(3 + Δ X)}^{2} - 4 (3 + Δ X) + 6 \\ = & (9 + 6 Δ X + Δ X^{2}) - 12 - 4 Δ X + 6 = Δ X^{2} + 2 Δ X + 3 \end{array}

$\begin{array}{l} Y_{2} & = & {(3 + Δ X)}^{2} - 4 (3 + Δ X) + 6 \\ = & (9 + 6 Δ X + Δ X^{2}) - 12 - 4 Δ X + 6 = Δ X^{2} + 2 Δ X + 3 \end{array}$

Thus,

Δ Y = Y_{2} - Y_{1} = (Δ X^{2} + 2 Δ X + 3) - 3 = Δ X^{2} + 2 Δ X

$Δ Y = Y_{2} - Y_{1} = (Δ X^{2} + 2 Δ X + 3) - 3 = Δ X^{2} + 2 Δ X$

The slope is then

b = \frac{Δ Y}{Δ X} = \frac{Δ X^{2} + 2 Δ X}{Δ X} = \frac{Δ X (Δ X + 2)}{Δ X} = Δ X + 2

$b = \frac{Δ Y}{Δ X} = \frac{Δ X^{2} + 2 Δ X}{Δ X} = \frac{Δ X (Δ X + 2)}{Δ X} = Δ X + 2$

As $Δ X$ $Δ X$ gets smaller, the value of $Δ Y/ Δ X$ $Δ Y/ Δ X$ approaches the slope of the tangent line. The value of the slope (b) in this example will approach 2. This is called the limit as $Δ X$ $Δ X$ approaches zero and is written as

lim_{Δ X \to 0} (Δ X + 2) = 2

$lim_{Δ X \to 0} (Δ X + 2) = 2$

It is obvious that at other points on the curve, the tangent line would have a different slope, as Y and $Δ Y$ $Δ Y$ would not be the values given here.

To find a general expression for the slope of the tangent line at any point on a curve, we can repeat this process for a general point X. Let $X_{1} = X$ $X_{1} = X$ and $X_{2} = X + Δ X,$ $X_{2} = X + Δ X,$ and let $Y_{1}$ $Y_{1}$ and $Y_{2}$ $Y_{2}$ represent the corresponding values for Y. For an equation of the form

Y = a X^{2} + b X + c

$Y = a X^{2} + b X + c$

we let

Y_{1} = a X^{2} + b X + c

$Y_{1} = a X^{2} + b X + c$

and

Y_{2} = a {(X + Δ X)}^{2} + b (X + Δ X) + c

$Y_{2} = a {(X + Δ X)}^{2} + b (X + Δ X) + c$

Expanding these expressions and simplifying, we find

Δ Y = Y_{2} - Y_{1} = b (Δ X) + 2 a X (Δ X) + c {(Δ X)}^{2}

$Δ Y = Y_{2} - Y_{1} = b (Δ X) + 2 a X (Δ X) + c {(Δ X)}^{2}$

Then

\frac{Δ Y}{Δ X} = \frac{b (Δ X) + 2 a X (Δ X) + c {(Δ X)}^{2}}{Δ X} = \frac{Δ X (b + 2 a X + c Δ X)}{Δ X} = b + 2 a X + c Δ X

$\frac{Δ Y}{Δ X} = \frac{b (Δ X) + 2 a X (Δ X) + c {(Δ X)}^{2}}{Δ X} = \frac{Δ X (b + 2 a X + c Δ X)}{Δ X} = b + 2 a X + c Δ X$

Taking the limit as $Δ$ $Δ$ X approaches zero, we have

lim_{Δ X \to 0} (b + 2 a X + c Δ X) = b + 2 a X

$lim_{Δ X \to 0} (b + 2 a X + c Δ X) = b + 2 a X$

This is the slope of the function at the point X, and it is called the derivative of Y. It is denoted as $Y'$ $Y'$ or $d Y / d X .$ $d Y / d X .$ The definition of a derivative is

Y' = \frac{d Y}{d X} = \underset{Δ X \to 0}{l i m} {\frac{Δ Y}{Δ X}}

$Y' = \frac{d Y}{d X} = \underset{Δ X \to 0}{l i m} {\frac{Δ Y}{Δ X}}$ (M6-3)

Fortunately, we will be using some common derivatives that are easy to remember, and it is not necessary to go through this process of finding the limit every time we wish to find a derivative.

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Table of Contents for M6.2 Slope of a Nonlinear Function

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Table of Contents for
M6.2 Slope of a Nonlinear Function