Question

there is a point on the graph of $y = \frac{x^{2}}{2}+1 - x$ where the normal is vertical. find the equation of this normal. give your answer in the form $x = a$ where $a$ is the value where the line crosses the $x$-axis.

Explanation:

Step1: Find the derivative

The function is \(y = \frac{x^{2}}{2}+1 - x\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(y^\prime=x - 1\).

Step2: Determine the slope of the normal

The slope of the tangent is \(m_{t}=y^\prime\). For a vertical normal, the slope of the tangent \(m_{t}=0\) (since the slope of a vertical line is undefined and \(m_{n}\times m_{t}=- 1\), when \(m_{n}\to\infty\), \(m_{t} = 0\)).
Set \(y^\prime=x - 1=0\), then \(x = 1\).

Step3: Find the \(y\) - coordinate

Substitute \(x = 1\) into the function \(y=\frac{x^{2}}{2}+1 - x\).
\(y=\frac{1^{2}}{2}+1 - 1=\frac{1}{2}\).

Step4: Write the equation of the normal

The normal is vertical, so its equation is \(x = 1\) (it crosses the \(x\) - axis at \(x = 1\)).

Answer:

\(x = 1\)