Find the equation of the tangent to the curve at \(x=2\).
Try the question yourself first. If you get stuck, open the hints before using the full walkthrough.
Start by finding the point of tangency, then differentiate to get the gradient, and finally use point-slope form.
Step 1
Find the point of tangency
At \(x=2\), what point on the curve does the tangent touch?
Step 2
Differentiate the function
Which derivative is correct?
Preview will appear here.
Step 3
Find the gradient at \(x=2\)
Substitute \(x=2\) into the derivative.
Preview will appear here.
Step 4
Use point-slope form
Which point-slope equation uses the correct point and gradient?
Step 5
Simplify the line equation
Rearrange the point-slope form into \(y=mx+c\), then type the equation.
Preview will appear here.
Step 6
Final answer and graph
The tangent has gradient \(1\) and passes through \(\left(2,\frac{16}{3}\right)\).
\[
y-\frac{16}{3}=x-2
\]
\[
y=x+\frac{10}{3}
\]
Great work. The tangent equation is \(y=x+\frac{10}{3}\).
Interactive tangent graph
This graph shows the cubic together with its tangent at \(x=2\). Use the controls to inspect the point of tangency and how the tangent sits against the curve.