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iGCSE (2021 Edition)

26.04 Equations of tangents (Extended)

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 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 $$y=f(x) the equation of the tangent at the point of contact $$(x1,y1) can be found using either:

  • $$y=mx+c  (gradient-intercept form)
  • $$yy1=m(xx1)   (point-gradient formula)

Where the gradient of the tangent is $$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 $$m=riserun. Then use the gradient and one of the points found in one of the forms above to find the equation. If it is clear on a graph a convenient point to use would be the $$y-intercept.

 

Finding the equation of the tangent to $$y=f(x) at $$x=a

Steps: The equation is of the form $$y=mx+c, we need to find $$m and then $$c.

  1. Find the gradient$$m, by evaluating $$m=f(a)
  2. Find the point of contact, the shared point between the function and the tangent by evaluating $$y=f(a). Giving us the point of contact as $$(a,f(a)). (If the point is given in the question, simply state it.)
  3. Find $$c by substituting the point of contact and gradient into the equation $$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) at the point pictured below.

Think: We can see the point of contact and the $$y-intercept clearly. Use these two points to find the gradient and then write the equation in the form $$y=mx+c.

Do: The point of contact is $$(1,3) and the $$y-intercept is $$(0,5). Thus, the gradient is:

$$m $$= $$y2y1x2x1
  $$= $$3(5)10
  $$= $$2

Since we have the $$y-intercept we know $$c=5, and hence the equation of the tangent is $$y=2x5.

Example 2

Find the equation of the tangent to $$f(x)=x2+5x3 at the point $$(2,11).

Think: We have been given the point of contact, $$(2,11), we also require the gradient of the tangent. So we will find the derivative and evaluate it at $$x=2. Then we can use both the gradient and point of contact to find the equation of the tangent.

Do:

$$f(x) $$= $$2x+5
Thus, $$m $$= $$f(2)
  $$= $$2(2)+5
  $$= $$9

We know the tangent has the form $$y=9x+c and goes through the point, $$(2,11). Substituting the point of contact in we can find $$c.

$$11 $$= $$9(2)+c
$$11 $$= $$18+c
$$c $$= $$7

Hence, the equation of the tangent to $$f(x) at $$x=2, is $$y=9x7.

 

Practice questions

question 1

Consider the curve $$f(x) drawn below along with $$g(x), which is a tangent to the curve.

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  1. What are the coordinates of the point at which $$g(x) is a tangent to the curve $$f(x)?

    Note that this point has integer coordinates. Give your answer in the form $$(a,b).

  2. What is the gradient of the tangent line?

  3. Hence determine the equation of the line $$y=g(x).

Question 2

Consider the parabola $$f(x)=x2+3x10.

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

  2. Determine the gradient of the tangent at the positive $$x-intercept.

 

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 3

The function $$f(x)=x3+ax2+bx+c has a $$y-intercept of $$3. The function has gradient of zero at $$x=1 and a root at $$x=3. Determine the values of $$a, $$band $$c.

Think: Break the information into parts and determine if the information given is about the function itself or its derivative.

  • The $$y-intercept tells us the original function passes through $$(0,3)
  • The gradient at $$x=1 tell us that the derivative equals zero when $$x=1, that is $$f(1)=0
  • The root tells us the original function passes through $$(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 $$y-intercept. Substituting $$(0,3) into the original function we get:

$$03+a×02+b×0+c $$= $$3
$$c $$= $$3


To use the information about the gradient, we will need to first find ourselves the derivative.

$$f(x)=3x2+2ax+b

Using the information $$f(1)=0, we obtain the equation:

$$3(1)2+2a(1)+b $$= $$0  
$$2a+b $$= $$3 ....Equation $$1

To use the information about the $$x-intercept, we can substitute $$(3,0) into the original function to obtain the equation:

$$(3)3+a(3)2+b(3)+3 $$= $$0  
$$27+9a3b+3 $$= $$0  
$$9a3b $$= $$24 ....Equation $$2

Solving equation $$1 and $$2 simultaneously (either with the elimination method, substitution method or with technology) we find that $$a=1 and $$b=5. Thus the original function was $$f(x)=x3+x25x+3.

 

Practice questions

Question 3

Consider the function $$f(x)=x2+5x.

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

  2. Hence state the coordinates of the point on the curve where the gradient is $$13.

Question 4

The curve $$y=ax3+bx2+2x17 has a gradient of $$58 at the point $$(2,31).

  1. Use the fact that the gradient of the curve at the point $$(2,31) is $$58 to express $$b in terms of $$a.

  2. Use the fact that the curve passes through the point $$(2,31) to express $$b in terms of $$a.

  3. Hence solve for $$a.

  4. Hence solve for $$b.

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