26.04 Equations of tangents (Extended)
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.
For a function $$y=f(x) the equation of the tangent at the point of contact $$(x1,y1) can be found using either:
- $$y=mx+c (gradient-intercept form)
- $$y−y1=m(x−x1) (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.
- Find the gradient, $$m, by evaluating $$m=f′(a)
- 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.)
- 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)
- 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:
| $$m | $$= | $$y2−y1x2−x1 |
| $$= | $$−3−(−5)1−0 | |
| $$= | $$2 |
Since we have the $$y-intercept we know $$c=−5, and hence the equation of the tangent is $$y=2x−5.
Example 2
Find the equation of the tangent to $$f(x)=x2+5x−3 at the point $$(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 $$y=9x−7.
Practice questions
question 1
Consider the curve $$f(x) drawn below along with $$g(x), which is a tangent to the curve.
What are the coordinates of the point at which $$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).
What is the gradient of the tangent line?
Hence determine the equation of the line $$y=g(x).
Question 2
Consider the parabola $$f(x)=x2+3x−10.
Solve for the $$x-intercepts. Write all solutions on the same line, separated by a comma.
Determine the gradient of the tangent at the positive $$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 | |
| $$−27+9a−3b+3 | $$= | $$0 | |
| $$9a−3b | $$= | $$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+x2−5x+3.
Practice questions
Question 3
Consider the function $$f(x)=x2+5x.
Find the $$x-coordinate of the point at which $$f(x) has a gradient of $$13.
Hence state the coordinates of the point on the curve where the gradient is $$13.
Question 4
The curve $$y=ax3+bx2+2x−17 has a gradient of $$58 at the point $$(2,31).
Use the fact that the gradient of the curve at the point $$(2,31) is $$58 to express $$b in terms of $$a.
Use the fact that the curve passes through the point $$(2,31) to express $$b in terms of $$a.
Hence solve for $$a.
Hence solve for $$b.