Sure, I'll guide you through the process of finding the equation of the normal to the function [tex]\( y = 2x \)[/tex] at the point [tex]\( (1, 2) \)[/tex].
1. Start with the given function:
[tex]\[
y = 2x
\][/tex]
2. Find the derivative of the function to determine the slope of the tangent line. For [tex]\( y = 2x \)[/tex], the derivative with respect to [tex]\( x \)[/tex] is:
[tex]\[
\frac{dy}{dx} = 2
\][/tex]
So, the slope of the tangent line at any point on the curve is [tex]\( 2 \)[/tex].
3. The point of interest is [tex]\( (1, 2) \)[/tex]. The slope of the tangent line at [tex]\( (1, 2) \)[/tex] is therefore [tex]\( 2 \)[/tex].
4. The slope of the normal line at a given point is the negative reciprocal of the slope of the tangent line. The negative reciprocal of [tex]\( 2 \)[/tex] is:
[tex]\[
-\frac{1}{2}
\][/tex]
Thus, the slope of the normal line at [tex]\( (1, 2) \)[/tex] is [tex]\( -\frac{1}{2} \)[/tex].
5. Use the point-slope form of the equation of a line to write the equation of the normal line. The point-slope form is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\( (x_1, y_1) \)[/tex] is a point on the line. Substituting [tex]\( m = -\frac{1}{2} \)[/tex] and [tex]\( (x_1, y_1) = (1, 2) \)[/tex] into the equation gives:
[tex]\[
y - 2 = -\frac{1}{2}(x - 1)
\][/tex]
6. Simplify the equation to standard form. Distribute the slope [tex]\( -\frac{1}{2} \)[/tex]:
[tex]\[
y - 2 = -\frac{1}{2}x + \frac{1}{2}
\][/tex]
7. Add [tex]\( 2 \)[/tex] to both sides to solve for [tex]\( y \)[/tex]:
[tex]\[
y = -\frac{1}{2}x + \frac{1}{2} + 2
\][/tex]
[tex]\[
y = -\frac{1}{2}x + \frac{5}{2}
\][/tex]
So, the equation of the normal to the curve [tex]\( y = 2x \)[/tex] at the point [tex]\( (1, 2) \)[/tex] is:
[tex]\[
y = -\frac{1}{2}x + \frac{5}{2}
\][/tex]
Or, equivalently, in standard form:
[tex]\[
y - 2 = -\frac{1}{2}(x - 1)
\][/tex]
