Question

Given a function f(x) where f'(2)=1/2 what is the equation of the line normal to the graph of the function at the point (2,-1)

Answer 1

let's reword all that jumbled material some

well, we know that the derivative of any function is just an equation to get the slope of a line at a given point, so once we have the derivative equation, we simply need either just one variable value or two assuming is implicit differentiation, but in the end the value we get is the slope.

we know that f'(2) = 1/2, the hell does that mean? well, it means the slope of a line on the curve when x = 2 is 1/2 for f(x), so we know that f(x) has a slope of m = 1/2 when x = 2.

a line at the point given, at (2 , -1) is a tangential line to the curve f(x), a normal to it, it's simply a perpendicular to it.

keeping in mind that perpendicular lines have negative reciprocal slopes,

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ \cfrac{1}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{1} \implies -2}}`

so when rambling on the line of the normal at (2,-1) we're really looking at the equation of a line whose slope is -2 and that it passes through (2 , -1)

`(\stackrel{x_1}{2}~,~\stackrel{y_1}{-1})\hspace{10em} \stackrel{slope}{m} ~=~ - 2 \\\\\\ \begin{array}c \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{- 2}(x-\stackrel{x_1}{2}) \implies y +1= -2 (x -2) \\\\\\ y+1=-2x+4\implies \boxed{y=-2x+3}`

Answer 2

Hello!

We are going to find the equation of the line that aligns with:

• f'(2)=1/2, which means that derivative of the function and plugging in 2 into "x" gives us an answer of 1/2.
• It passes through the point (2,-1) in the graph

In order to find which equation works with our given information, we will use the approach of getting the derivative of the equation and plug in "2" to see if it equals to 1/2.

Solve:

Solve each equation by getting the derivative and plugging in "2" to "x" if applicable.

When getting the derivative of each equation, we'll use the power rule: nx^(n-1).

Also, know that the derivative of a constant is 0.

A).

y = -2x + 3

Calculate the derivative

y' = -2

We won't be able to plug in "2" since there is no "x" variable, and the derivative function doesn't equal 1/2, which makes this equation wrong in this question.

B).

y = -1/2x

Calculate the derivative

y' = -1/2

Like answer choice "A", we won't be able to plug in "2" since there is no "x" variable, and the derivative function doesn't equal 1/2, which makes this equation wrong in this question.

C).

y = -1

Calculate the derivative

y' = 0

Like our above answer choices, we won't be able to plug in "2" since there is no "x" variable, and the derivative function doesn't equal 1/2, which makes this equation wrong in this question.

D).

y = 1/2x - 2

Calculate the derivative

y' = 1/2

Even though we can't plug in "2" to "x" since it doesn't appear in the derivative function, this answer would be correct since the derivative function equals 1/2. This would be our answer.

Answer:

D). y = 1/2x - 2