# Secant and Tangent Lines

In order to find the gradient of tangent of a polynomial, we have to use the first order differentiation, i.e. \(f'(x)\) (refer to differentiation).

#### Contents

## Introduction

In order to find the average rate of change of a function on the interval \( [a, b]\), you might draw a **secant line** passing through \( (a, f(a) ) \) and \( (b, f(b)) \). The average rate of change on that interval is the slope of the secant line, or \( \dfrac{f(b)-f(a)}{b-a} \).

However, what if we wanted to find instantaneous rate of change of a function, let's say at \( (c, f(c)) \)? We would draw a **tangent line** through \( (c, f(c)) \) and only \( (c, f(c)) \). Then we would try to use the previous formula to find the slope of the tangent line, which would give us the instantaneous rate of change, and say \( \dfrac{f(c)-f(c)}{c-c} = \dfrac{0}{0} \). Uh oh! We can't have \( \dfrac{0}{0} \)! So we have to find a different approach.

What we could do is have a second point \( (c+h, f(c+h )) \), where \( h\) is a very small number, approaching zero. Then, we can find what the slope **approaches** as the secant line eventually becomes a tangent line. We write this as \( \displaystyle \lim_{h \to 0} \dfrac{f(c+h)-f(c)}{h} \). If we wanted to find the slope of tangent at any \(x \) value, we could write \( \displaystyle \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \), commonly known as the **difference quotient**.

## Finding slope

To find the slope \(m\) of a curve at a particular point, we differentiate the equation of the curve. If the given curve is \(y=f(x),\) we evaluate \(\dfrac { dy }{ dx } \) or \(f'(x)\) and substitute the value of \(x\) to find the slope.

For a line of the form (or any other form) \(y=mx+c,\) we can find its slope by simply taking any two values of \(x,\) \({x}_{1}\) and \({x}_{2},\) and their respective \( y\) values, \({y}_{1}\) and \({y}_{2}\). We find the slope by the formula \(\tan \theta =\dfrac {{y}_{1}-{y}_{2}}{{x}_{1}- {x}_{2}}\). In the case of curves our approach is somewhat different. In the above case, we had \( \Delta y={y}_{1}-{y}_{2} \) and \(\Delta x={x}_{1}-{x}_{2} \). Now we need to find the slope of tangent to a curve at some point. To do this we again need

\[\tan \theta =\frac {{y}_{1}-{y}_{2}}{{x}_{1}- {x}_{2}},\]

but this time \( \Delta y\) and \( \Delta x\) tend to zero, which means the interval is very small because it is a tangent at a point.

Notice that as the colored pairs of \(x_1\) and \(x_2\) come closer, the tangent shifts to a point on the graph.

When this happens we replace

\[ \frac { \Delta y }{ \Delta x } \quad by\quad \frac { dy }{ dx } \]

and therefore find \(\frac { dy }{ dx }.\) \(_\square\)

## What is the slope of curve \(y=x^4-x^3\) at \(x=1?\)

The given curve is \(y=x^4-x^3.\) Evaluating \(\frac { dy }{ dx },\) we have

\[\frac{dy}{dx}=4x^3-3x^2.\]

Substituting \(x=1\) into this gives

\[ \frac{dy}{dx}\Big \rvert _{x=1}=4-3=1.\]

Therefore the slope of the given curve at \(x=1\) is \(1.\) \( _\square \)

## If the curve \(y=2x^3-bx+a\) passes through \((19, 2)\) and its slope at \(x=1\) is \(5,\) then what are the values of \(a\) and \(b?\)

The given curve is \(y=2x^3-bx+a.\) Evaluating \(\frac { dy }{ dx } \), we have

\[\frac{dy}{dx}=6x^2-b.\]

Substituting \(x=1\) gives

\[\frac{dy}{dx}=6-b.\]

Since the slope of the curve at \(x=1\) is \(5,\) we have

\[6-b=5 \Rightarrow b=1.\]

Now, substituting \((19, 2)\) into \(y=2x^3-bx+a\) gives

\[2=2(19^3)-19+a \Rightarrow a=-13697.\]

Therefore, \(a=-13697\) and \(b=1.\) \( _\square \)

## Slope of a Curve

See main article: slope of a curve

For a problem of the slope of a curve, you should have a polynomial equation. The question would require you to find the tangent line at a point with the coordinates of that point provided. After finding the gradient of the tangent, you can substitute the values of the gradient and coordinates of the tangent line into the general linear equation: \(y=mx+c.\) You will then be able to find what the value of \(c\) is, and then you have found the tangent line at a point.

You can also use the formula. The formula for the line tangent to \(f(x)\) at point \((a,f(a))\) is

\[y-f(a)=f'(a)(x-a).\]

## Find the tangent line at \(x=3\) of the function \(f(x) = x^2 - 2x + 3.\)

First, find the derivative of \(f(x):\) \(f'(x) = 2x - 2.\)

At \(x=3,\) we have \(f'(x) = 4.\) (This is the gradient of the tangent at \(x=3\).)

By finding the \(y\) (or \(f(x)\)) value when \(x=3,\) we have \(y=6.\)

Substituting into the general linear equation, \(6=3(4)+c \Rightarrow c=-6.\)

Therefore, the equation of the tangent line is \(y=4x-6.\) \(_\square\)

**Cite as:**Secant and Tangent Lines.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/tangent-line-point/