there is a point on the graph of $y = \\frac{x^{2}}{2}+1 - x$ where the normal is vertical. find the…

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.

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.

Answer

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)