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AustraliaVIC
VCE 11 Methods 2023

10.02 Tangents and their equations

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). 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.)
  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:  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 $=$= $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1
  $=$= $\frac{-3-\left(-5\right)}{1-0}$3(5)10
  $=$= $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=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: 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=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 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}}$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 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.

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 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+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) 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+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.

 

Outcomes

U1.AoS3.3

use of gradient of a tangent at a point on the graph of a function to describe and measure instantaneous rate of change of the function, including consideration of where the rate of change is positive, negative or zero, and the relationship of the gradient function to features of the graph of the original function.

U1.AoS3.5

use graphical, numerical and algebraic approaches to find an approximate value or the exact value (as appropriate) for the gradient of a secant or tangent to a curve at a given point

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