topic badge
Standard level

10.07 Tangents

Lesson

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.

Equation of tangents

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)
  • $y-y_1=m\left(x-x_1\right)$yy1=m(xx1)   (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.

  1. Find the gradient$m$m, by evaluating $m=f'(a)$m=f(a)
  2. 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.)
  3. 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)
  4. 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 $=$= $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1
  $=$= $\frac{-3-\left(-5\right)}{1-0}$3(5)10
  $=$= $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=2x5.

Example 2

Find the equation of the tangent to $f(x)=x^2+5x-3$f(x)=x2+5x3 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=9x7.

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}}$12x12
  $=$= $\frac{1}{2\sqrt{x}}$12x
$\therefore f'(4)$f(4) $=$= $\frac{1}{2\sqrt{4}}$124
  $=$= $\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.

Loading Graph...

  1. 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).

  2. What is the gradient of the tangent line?

  3. 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+3x10.

  1. Solve for the $x$x-intercepts. Write all solutions on the same line, separated by a comma.

  2. 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)=5x at the point $\left(\frac{1}{9},\frac{5}{3}\right)$(19,53).

  1. Firstly, find the gradient of the function $f\left(x\right)=5\sqrt{x}$f(x)=5x at $x=\frac{1}{9}$x=19.

  2. Hence find the equation of the tangent to the curve $f\left(x\right)=5\sqrt{x}$f(x)=5x 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+9a3b+3 $=$= $0$0  
$9a-3b$9a3b $=$= $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+x25x+3.

 

Practice questions

Question 4

Consider the function $f\left(x\right)=x^2+5x$f(x)=x2+5x.

  1. Find the $x$x-coordinate of the point at which $f\left(x\right)$f(x) has a gradient of $13$13.

  2. 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+2x17 has a gradient of $58$58 at the point $\left(2,31\right)$(2,31).

  1. 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.

  2. 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.

  3. Hence solve for $a$a.

  4. Hence solve for $b$b.

What is Mathspace

About Mathspace