Question
The graph below shows the function \(y=\sqrt{x+2}\), and the normal to the function at the point where the function intersects the \(y\)-axis.
Find the coordinates of point \(P\), the \(x\)-intercept of the normal.
Try the question yourself first. If you get stuck, open the hints before using the full walkthrough.
You must use calculus and show any derivatives that you need.
Step 1
Find the point on the curve when \(x=0\)
The function meets the \(y\)-axis where \(x=0\).
Step 2
Rewrite the square root as a power
This makes differentiation easier.
Step 3
Differentiate using the chain rule
What is the derivative of \(y=\sqrt{x+2}\)?
Step 4
Find the tangent gradient at \(x=0\)
Substitute \(x=0\) into the derivative.
Step 5
Find the normal gradient
The gradient of the normal is the negative reciprocal of the tangent gradient.
Step 6
Find the x-intercept of the normal
Use the point \((0,\sqrt{2})\) and gradient \(-2\sqrt{2}\) in the line equation.
\[
y-\sqrt{2}=-2\sqrt{2}(x-0)
\]
\[
y=-2\sqrt{2}x+\sqrt{2}
\]
When the line crosses the \(x\)-axis, \(y=0\). What is \(x\)?
Step 7
Final answer and graph
Point \(P\), the \(x\)-intercept of the normal, is
\[
P=\left(\frac{1}{2},0\right)
\]