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Find the equation of the normal to the curve
y=(x-2)/(2x+1) at the point where the curve cuts the
x-axis.

Find the equation of the normal to the curve $y=\frac{x-2}{2 x+1}$ at the point where the curve cuts the $x$ -axis.

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Find equations of tangent lines using limits

Full solution

Q. Find the equation of the normal to the curve

y=\frac{x-2}{2 x+1}

at the point where the curve cuts the

x

-axis.

Find x-intercept: First, we need to find the point where the curve cuts the x-axis. This happens when $y=0$ . $\newline$ $y = \frac{x-2}{2x+1} = 0$ $\newline$ To find the x-coordinate of this point, we set the numerator of the fraction to zero because a fraction is zero if and only if its numerator is zero. $\newline$ $x - 2 = 0$ $\newline$ $x = 2$
Determine point coordinates: Now we have the $x$ -coordinate of the point where the curve cuts the $x$ -axis. The $y$ -coordinate is already known to be $0$ since it's on the $x$ -axis. So the point is $(2, 0)$ .
Calculate tangent slope: Next, we need to find the slope of the tangent to the curve at the point $(2, 0)$ . To do this, we find the derivative of $y$ with respect to $x$ . $y = \frac{x-2}{2x+1}$ Using the quotient rule, the derivative $y'$ is: $y' = \frac{(2x+1)(1) - (x-2)(2)}{(2x+1)^2}$ $y' = \frac{2x+1 - 2x + 4}{(2x+1)^2}$ $y' = \frac{5}{(2x+1)^2}$
Evaluate slope at $x=2$ : Now we evaluate the derivative at the point $x=2$ to find the slope of the tangent. $\newline$ $y'(2) = \frac{5}{(2\cdot2+1)^2}$ $\newline$ $y'(2) = \frac{5}{(4+1)^2}$ $\newline$ $y'(2) = \frac{5}{25}$ $\newline$ $y'(2) = \frac{1}{5}$
Find normal slope: The slope of the normal is the negative reciprocal of the slope of the tangent. Since the slope of the tangent at $x=2$ is $\frac{1}{5}$ , the slope of the normal is $-5$ .
Derive normal equation: Now we have the slope of the normal and a point through which it passes. We can use the point-slope form of the equation of a line to find the equation of the normal. $\newline$ The point-slope form is: $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is the point. $\newline$ Using the point $(2, 0)$ and the slope $-5$ , the equation is: $\newline$ $y - 0 = -5(x - 2)$ $\newline$ $y = -5x + 10$

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