A tangent to a function is a straight line and as such we can use our knowledge of linear functions to find the equation of a tangent. Our new technique of differentiation will allow us to find the gradient of the tangent to a function at any given point.
For a function $y=f(x)$y=f(x) the equation of the tangent at the point of contact $(x_1,y_1)$(x1,y1) can be found using either:
- $y=mx+c$y=mx+c (gradient-intercept form)
- $y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1) (point-gradient formula)
Where the gradient of the tangent is $m=f'(x_1)$m=f′(x1).
Finding the equation of a tangent from a graph
From a graph, look for two easily identifiable points then calculate the gradient using $m=\frac{rise}{run}$m=riserun. Then, use the gradient and one of the points found in one of the forms above to find the equation. If it is clear on a graph a convenient point to use would be the $y$y-intercept.
Finding the equation of the tangent to $y=f(x)$y=f(x) at $x=a$x=a
Steps: The equation is of the form $y=mx+c$y=mx+c, we need to find $m$m and then $c$c.
- Find the gradient, $m$m, by evaluating $m=f'(a)$m=f′(a)
- Find the point of contact, the shared point between the function and the tangent by evaluating $y=f(a)$y=f(a). Give the point of contact as $\left(a,f(a)\right)$(a,f(a)). (If the point is given in the question, simply state it.)
- Find $c$c by substituting the point of contact and gradient into the equation $y=mx+c$y=mx+c and then rearrange. (Or use the point gradient form of the equation)
- State the equation of the tangent.
Worked examples
Example 1
Find the equation of the tangent to $f(x)$f(x) at the point pictured below.
Think: The point of contact and the $y$y-intercept can be seen clearly. Use these two points to find the gradient and then write the equation in the form $y=mx+c$y=mx+c.
Do: The point of contact is $\left(1,-3\right)$(1,−3) and the $y$y-intercept is $\left(0,-5\right)$(0,−5). Thus, the gradient is:
| $m$m | $=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |
| $=$= | $\frac{-3-\left(-5\right)}{1-0}$−3−(−5)1−0 | |
| $=$= | $2$2 |
Since the $y$y-intercept is shown it is known that $c=-5$c=−5, and hence the equation of the tangent is $y=2x-5$y=2x−5.
Example 2
Find the equation of the tangent to $f(x)=x^2+5x-3$f(x)=x2+5x−3 at the point $\left(2,11\right)$(2,11).
Think: Since the the point of contact has been given , $\left(2,11\right)$(2,11), the gradient of the tangent can be found. So find the derivative and evaluate it at $x=2$x=2. Then use both the gradient and point of contact to find the equation of the tangent.
Do:
| $f'(x)$f′(x) | $=$= | $2x+5$2x+5 |
| Thus, $m$m | $=$= | $f'(2)$f′(2) |
| $=$= | $2(2)+5$2(2)+5 | |
| $=$= | $9$9 |
The tangent has the form $y=9x+c$y=9x+c and goes through the point, $\left(2,11\right)$(2,11). By substituting the point of contact in, $c$ccan be found.
| $11$11 | $=$= | $9\left(2\right)+c$9(2)+c |
| $11$11 | $=$= | $18+c$18+c |
| $\therefore c$∴c | $=$= | $-7$−7 |
Hence, the equation of the tangent to $f(x)$f(x) at $x=2$x=2, is $y=9x-7$y=9x−7.
Example 3
Find the equation of the tangent to $f(x)=\sqrt{x}$f(x)=√x at $x=4$x=4.
Think: This time the point of contact has not been given but we can evaluate the function at $x=4$x=4 to find it. It is also necessary to find the derivative in order to determine the gradient of the tangent.
Do:
Find the gradient of tangent:
| $f(x)$f(x) | $=$= | $x^{\frac{1}{2}}$x12 |
| $f'(x)$f′(x) | $=$= | $\frac{1}{2}x^{-\frac{1}{2}}$12x−12 |
| $=$= | $\frac{1}{2\sqrt{x}}$12√x | |
| $\therefore f'(4)$∴f′(4) | $=$= | $\frac{1}{2\sqrt{4}}$12√4 |
| $=$= | $\frac{1}{4}$14 |
Find the point of contact, when $x=4$x=4:
| $f(4)$f(4) | $=$= | $\sqrt{4}$√4 |
| $=$= | $2$2 |
Thus, the point of contact is $(4,2)$(4,2).
Find $c$c:
The tangent is of the form $y=\frac{1}{4}x+c$y=14x+c and passes through $(4,2)$(4,2). Substituting into the equation and get:
| $2$2 | $=$= | $\frac{1}{4}\left(4\right)+c$14(4)+c |
| $2$2 | $=$= | $1+c$1+c |
| $\therefore c$∴c | $=$= | $1$1 |
Hence, the equation of the tangent to $f(x)$f(x) at $x=4$x=4 is $y=\frac{1}{4}x+1$y=14x+1.
Practice questions
question 1
Consider the curve $f\left(x\right)$f(x) drawn below along with $g\left(x\right)$g(x), which is a tangent to the curve.
What are the coordinates of the point at which $g\left(x\right)$g(x) is a tangent to the curve $f\left(x\right)$f(x)?
Note that this point has integer coordinates. Give your answer in the form $\left(a,b\right)$(a,b).
What is the gradient of the tangent line?
Hence determine the equation of the line $y=g\left(x\right)$y=g(x).
Question 2
Consider the parabola $f\left(x\right)=x^2+3x-10$f(x)=x2+3x−10.
Solve for the $x$x-intercepts. Write all solutions on the same line, separated by a comma.
Determine the gradient of the tangent at the positive $x$x-intercept.
Question 3
Consider the tangent to the curve $f\left(x\right)=5\sqrt{x}$f(x)=5√x at the point $\left(\frac{1}{9},\frac{5}{3}\right)$(19,53).
Firstly, find the gradient of the function $f\left(x\right)=5\sqrt{x}$f(x)=5√x at $x=\frac{1}{9}$x=19.
Hence find the equation of the tangent to the curve $f\left(x\right)=5\sqrt{x}$f(x)=5√x at the point $\left(\frac{1}{9},\frac{5}{3}\right)$(19,53).
Express the equation of the tangent line in the form $y=mx+c$y=mx+c.
Problem-solving using gradient information
Common problem-solving questions include using the derivative to determine the point(s) on a curve where a given gradient occurs or using the information about gradients and the original function to determine unknown coefficients in the original function.
Worked example
Example 4
The function $f\left(x\right)=x^3+ax^2+bx+c$f(x)=x3+ax2+bx+c has a $y$y-intercept of $3$3. The function has gradient of zero at $x=1$x=1 and a root at $x=-3$x=−3. Determine the values of $a$a, $b$band $c$c.
Think: Break the information into parts and determine if the information given is about the function itself or its derivative.
- The $y$y-intercept tells shows that the original function passes through $\left(0,3\right)$(0,3)
- The gradient at $x=1$x=1 shows that the derivative equals zero when $x=1$x=1, that is $f'(1)=0$f′(1)=0
- The root shows that the original function passes through $\left(-3,0\right)$(−3,0)
Three pieces of information and three unknowns are seen, so by using simultaneous equations the solutions to a, b and c can be found.
Do: Begin by using the information about the $y$y-intercept. Substituting $\left(0,3\right)$(0,3) into the original function we get:
| $0^3+a\times0^2+b\times0+c$03+a×02+b×0+c | $=$= | $3$3 |
| $\therefore c$∴c | $=$= | $3$3 |
To use the information about the gradient, first find the derivative.
$f'\left(x\right)=3x^2+2ax+b$f′(x)=3x2+2ax+b
Using the information $f'(1)=0$f′(1)=0, the following equation can be found:
| $3(1)^2+2a(1)+b$3(1)2+2a(1)+b | $=$= | $0$0 | |
| $2a+b$2a+b | $=$= | $-3$−3 | ....Equation $1$1 |
To use the information about the $x$x-intercept, substitute $\left(-3,0\right)$(−3,0) into the original function to obtain the equation:
| $\left(-3\right)^3+a\left(-3\right)^2+b\left(-3\right)+3$(−3)3+a(−3)2+b(−3)+3 | $=$= | $0$0 | |
| $-27+9a-3b+3$−27+9a−3b+3 | $=$= | $0$0 | |
| $9a-3b$9a−3b | $=$= | $24$24 | ....Equation $2$2 |
Solving equation $1$1 and $2$2 simultaneously (either with the elimination method, substitution method or with technology) it is found that $a=1$a=1 and $b=-5$b=−5. Thus the original function is $f(x)=x^3+x^2-5x+3$f(x)=x3+x2−5x+3.
Practice questions
Question 4
Consider the function $f\left(x\right)=x^2+5x$f(x)=x2+5x.
Find the $x$x-coordinate of the point at which $f\left(x\right)$f(x) has a gradient of $13$13.
Hence state the coordinates of the point on the curve where the gradient is $13$13.
Question 5
The curve $y=ax^3+bx^2+2x-17$y=ax3+bx2+2x−17 has a gradient of $58$58 at the point $\left(2,31\right)$(2,31).
Use the fact that the gradient of the curve at the point $\left(2,31\right)$(2,31) is $58$58 to express $b$b in terms of $a$a.
Use the fact that the curve passes through the point $\left(2,31\right)$(2,31) to express $b$b in terms of $a$a.
Hence solve for $a$a.
Hence solve for $b$b.
The tangent function and Newton's Method
Being able to find the tangent at a point has many applications. One of these is approximating the solutions (or roots) of a function.
Suppose an approximate solution $f(x)=0$f(x)=0 for some function $f(x)$f(x)is needed. Suppose some approximate idea of where the solution is is known - call this initial approximation $x_0$x0. Often, this approximation may be found by sketching a graph and estimating a “close” value to the solution from the sketch. If an interval is provided, the guess may also be made by bisecting the interval (choosing the mid-point).
To find an approximation for the solution, recall that the point gradient form of the equation to the tangent at a point $x_0$x0 is given by
$y-y_0=m\left(x-x_0\right)$y−y0=m(x−x0)
where $m=f'(x_0)$m=f′(x0) and $y_0$y0 is the value of the function at $x=x_0$x=x0.
If $f(x)=y$f(x)=y, this equation can be re-arranged and written in function notation:
$y=f(x_0)+f'(x_0)(x-x_0)$y=f(x0)+f′(x0)(x−x0)
for some point $(x,y)$(x,y). This tangent will cross the x-axis at a point which we will call $x_1$x1. This is demonstrated in the following image:
Notice that the initial point $x_0$x0 that was chosen is not very close to the solution where $f(x)=0$f(x)=0, but the x-intercept of the tangent at $x_1$x1 is a little closer. If this process is repeated many times, the point would get closer and closer to the actual solution.
The point $x_1$x1 can be expressed in terms of $x_0$x0 by substituting the point $(x_1,0)$(x1,0) into the tangent equation:
$0=f(x_0)+f'(x_0)(x_1-x_0)$0=f(x0)+f′(x0)(x1−x0)
Solving for $x_1$x1 gives:
$-f'(x_0)(x_1-x_0)=f(x_0)$−f′(x0)(x1−x0)=f(x0)
$x_1-x_0=-\frac{f(x_0)}{f'(x_0)}$x1−x0=−f(x0)f′(x0)
$x_1=x_0-\frac{f(x_0)}{f'(x_0)}$x1=x0−f(x0)f′(x0)
Repeating this process again would give a point $x_2$x2 which would sit on the new tangent line and would move closer to the actual solution. Using the same reasoning, the value for $x_2$x2 would be given by:
$x_2=x_1-\frac{f(x_1)}{f'(x_1)}$x2=x1−f(x1)f′(x1)
This process could be repeated as many times as necessary to get an approximation that is sufficiently “close enough” to the required solution. This process is called Newton’s Method, after the mathematician and physicist Sir Isaac Newton.
Newton’s Method can be generalised as follows:
If $x_n$xn is the approximation of a solution where $f(x)=0$f(x)=0, and if the derivative $f'(x_n)\ne0$f′(xn)≠0, then the next approximation is given by
$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$xn+1=xn−f(xn)f′(xn) for $n=0,1,2,\dots$n=0,1,2,…
Note that this process is recursive that is, it is a repeated procedure used to generate successive results. For this reason, Newton's Method is very well suited to the use of technology.
Experiment with the following applet to see Newton's Method in action:
To decide when to stop with an approximation that is “good enough”, it is common to first decide on a number of decimal places to which two successive solutions agree. That is, when two successive solutions are the same when rounded to that number of decimal places, this would be the final approximation.
This process is explored in the following example.
Worked example
EXAMPLE 5
Use Newton’s Method to determine an approximation to the solution to $f(x)=2x^2-4$f(x)=2x2−4 that lies in the interval $[1,2]$[1,2]. Find the approximation to four decimal places.
Think: To make an initial approximation, sketch the graph:
From this, an initial guess of $x_0=1.5$x0=1.5 would be reasonable as it is the mid-point of the given interval and tangents drawn from the right side of the solutions will move closer to the actual solution.
Do: Remember that the purpose of this method is finding the solution to the equation $f(x)=0$f(x)=0. So here, that is $2x^2-4=0$2x2−4=0. While this is easy enough to solve algebraically, Newton's Method will be used to demonstrate the process.
Write the general formula for Newton's Method with the given $f(x)=2x^2-4$f(x)=2x2−4 and where $f'(x)=4x$f′(x)=4x:
$x_{n+1}=x_n-\frac{2x^2-4}{4x}$xn+1=xn−2x2−44x
The first approximation, with $x_0=1.5$x0=1.5, will be:
$x_1=1.5-\frac{2(1.5)^2-4}{4(1.5)}=1.416667$x1=1.5−2(1.5)2−44(1.5)=1.416667
The next approximation would be:
$x_2=1.416667-\frac{2(1.416667)^2-4}{4(1.416667)}=1.414216$x2=1.416667−2(1.416667)2−44(1.416667)=1.414216
Continuing:
$x_3=1.414216-\frac{2(1.414216)^2-4}{4(1.414216)}=1.414213$x3=1.414216−2(1.414216)2−44(1.414216)=1.414213
Notice that, to four decimal places, the solutions $x_2=1.4142$x2=1.4142 and $x_3=1.4142$x3=1.4142. Since two successive solutions agree to four decimal places, the process ends. The approximate solution of $f(x)=0$f(x)=0 to four decimal places is $x_3=1.4142$x3=1.4142.
Check (optional): Since this is a simple quadratic, solving $f(x)=0$f(x)=0 gives $2x^2-4=0$2x2−4=0 which leads to the positive solution $x=\sqrt{2}=1.4142$x=√2=1.4142 to four decimal places, which agrees with the approximation found.
Practice question
QUESTION 6
Consider the function $f\left(x\right)=x^4-x^3+x^2-3$f(x)=x4−x3+x2−3.
Find $f'\left(x\right)$f′(x).
Use three iterations of Newton's method to approximate a solution to $f\left(x\right)=0$f(x)=0, starting with $x=1$x=1.
Round each answer to four decimal places.
$n$n $x_n$xn $x_{n+1}$xn+1 $0$0 $1$1 $\editable{}$ $1$1 $\editable{}$ $\editable{}$ $2$2 $\editable{}$ $\editable{}$ Hence state the approximate solution to $f\left(x\right)=0$f(x)=0.
Give your answer to four decimal places.