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India
Class XI

Equations of Tangent Lines

Lesson

Now that we can find the gradient of a tangent line, we are going to be able to find the equation of a tangent line. 

The following lessons were covered way back when we were learning how to find the gradient of a straight line, well; because tangents are straight lines all the strategies we learnt then still apply now.

  • Gradient-Intercept form (uses the gradient and the y intercept to form the equation) 
  • Point-Gradient formula (uses the gradient and any other point on the line to form the equation) 
  • The two point formula (uses any two points on the line to form the equation) 
  • Finding the equation of a line (a mixed set to practice all three forms) 

 

Examples

Example 1

Find the equation of the tangent to the curve, at the point $(1,-3)$(1,3).

We have the point $(1,-3)$(1,3) and we can easily identify up to $3$3 other points, the points $(3,1),(2,-1)$(3,1),(2,1) and $(0,-5)$(0,5) are also on the tangent line.  We need only choose one of them to help us find the equation.  I'm going to choose $(0,-5)$(0,5).

Now that we have $2$2 points on the line $(1,-3)$(1,3) and $(0.-5)$(0.5) I can use the two point formula to construct the equation.

$\frac{y-y_1}{x-x_1}$yy1xx1 $=$= $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1
$\frac{y--3}{x-1}$y3x1 $=$= $\frac{-5--3}{0-1}$5301
$\frac{y+3}{x-1}$y+3x1 $=$= $\frac{-2}{-1}$21
$y+3$y+3 $=$= $2(x-1)$2(x1)
$y+3$y+3 $=$= $2x-2$2x2
$y$y $=$= $2x-5$2x5

 

For this question, I could also have used the point gradient formula by first finding the gradient ($m=2$m=2) and any of the points listed above. I could also have used the gradient intercept formula because we have the y intercept of $-5$5.  

Worked Examples:

question 1

Answer the following.

  1. Consider the function $f\left(x\right)$f(x) drawn below.

    Plot the function $g\left(x\right)=2x-1$g(x)=2x1 on the graph.

    Loading Graph...
  2. Is $g\left(x\right)$g(x) a tangent to $f\left(x\right)$f(x)?

    Yes

    A

    No

    B

question 2

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 3

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).

  4. What is the $x$x-coordinate of the point on the curve at which we could draw a tangent line that has the same gradient as $g\left(x\right)$g(x)?

 

Outcomes

11.C.LD.1

Derivative introduced as rate of change both as that of distance function and geometrically, intuitive idea of limit. Definition of derivative, relate it to slope of tangent of the curve, derivative of sum, difference, product and quotient of functions. Derivatives of polynomial and trigonometric functions.

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