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). Giving us 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: We can see the point of contact and the $y$y-intercept 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 we have the $y$y-intercept we know $c=-5$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: We have been given the point of contact, $\left(2,11\right)$(2,11), we also require the gradient of the tangent. So we will find the derivative and evaluate it at $x=2$x=2. Then we can 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 |
We know the tangent has the form $y=9x+c$y=9x+c and goes through the point, $\left(2,11\right)$(2,11). Substituting the point of contact in we can find $c$c.
| $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 we have not been given the point of contact but we can evaluate the function at $x=4$x=4 to find it. We also need to find the derivative to find 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 we 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 us the original function passes through $\left(0,3\right)$(0,3)
- The gradient at $x=1$x=1 tell us that the derivative equals zero when $x=1$x=1, that is $f'(1)=0$f′(1)=0
- The root tells us the original function passes through $\left(-3,0\right)$(−3,0)
We have three pieces of information and three unknowns, so we should be able to solve using simultaneous equations.
Do: We can 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, we will need to first find ourselves the derivative.
$f'\left(x\right)=3x^2+2ax+b$f′(x)=3x2+2ax+b
Using the information $f'(1)=0$f′(1)=0, we obtain the equation:
| $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, we can 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) we find that $a=1$a=1 and $b=-5$b=−5. Thus the original function was $f(x)=x^3+x^2-5x+3$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.