Rates of Change

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)

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−y1x−x1 |
$=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−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.

Answer the following.

Consider the function $f\left(x\right)$

`f`(`x`) drawn below.Plot the function $g\left(x\right)=2x-1$

`g`(`x`)=2`x`−1 on the graph.Loading Graph...Is $g\left(x\right)$

`g`(`x`) a tangent to $f\left(x\right)$`f`(`x`)?Yes

ANo

BYes

ANo

B

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...

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`).What is the gradient of the tangent line?

Hence determine the equation of the line $y=g\left(x\right)$

`y`=`g`(`x`).

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...

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`).What is the gradient of the tangent line?

Hence determine the equation of the line $y=g\left(x\right)$

`y`=`g`(`x`).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`)?

Apply differentiation and anti-differentiation techniques to polynomials

Apply calculus methods in solving problems