 New Zealand
Level 7 - NCEA Level 2

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.

 $\frac{y-y_1}{x-x_1}$y−y1​x−x1​​ $=$= $\frac{y_2-y_1}{x_2-x_1}$y2​−y1​x2​−x1​​ $\frac{y--3}{x-1}$y−−3x−1​ $=$= $\frac{-5--3}{0-1}$−5−−30−1​ $\frac{y+3}{x-1}$y+3x−1​ $=$= $\frac{-2}{-1}$−2−1​ $y+3$y+3 $=$= $2(x-1)$2(x−1) $y+3$y+3 $=$= $2x-2$2x−2 $y$y $=$= $2x-5$2x−5

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

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.

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

Yes

A

No

B

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.

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.

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

M7-10

Apply differentiation and anti-differentiation techniques to polynomials

91262

Apply calculus methods in solving problems