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.
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)=2x−1 on the graph.
Is $g\left(x\right)$g(x) a tangent to $f\left(x\right)$f(x)?
Yes
No
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.
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.
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)?