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Consider the curve given by the equation
3x^(2)+y^(4)+6x=253. It can be shown that

(dy)/(dx)=(-6(x+1))/(4y^(3))
Write the equation of the horizontal line that is tangent to the curve and is above the
x-axis.

Consider the curve given by the equation $3 x^{2}+y^{4}+6 x=253$ . It can be shown that $\frac{d y}{d x}=\frac{-6(x+1)}{4 y^{3}}.$ $\newline$ Write the equation of the horizontal line that is tangent to the curve and is above the $x$ -axis.

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Find equations of tangent lines using limits

Full solution

Q. Consider the curve given by the equation

3 x^{2}+y^{4}+6 x=253

. It can be shown that

\frac{d y}{d x}=\frac{-6(x+1)}{4 y^{3}}.

\newline

Write the equation of the horizontal line that is tangent to the curve and is above the

x

-axis.

Find Tangent Point: To find the equation of a horizontal line that is tangent to the curve, we need to find a point on the curve where the derivative $\frac{dy}{dx}$ is equal to $0$ , because the slope of a horizontal line is $0$ .
Derivative Calculation: The derivative $\frac{dy}{dx}$ is given as $\frac{-6(x+1)}{4y^3}$ . For a horizontal tangent line, $\frac{dy}{dx}$ must be equal to $0$ . Therefore, we set the numerator of the derivative equal to zero to find the $x$ -value of the point of tangency. $\newline$ $-6(x+1) = 0$
Solve for x: Solving for x, we get: $\newline$ $-6(x+1) = 0$ $\newline$ $-6x - 6 = 0$ $\newline$ $-6x = 6$ $\newline$ $x = -1$
Find y-Value: Now that we have the x-value, we need to find the corresponding y-value(s) on the curve. We substitute $x = -1$ into the original equation of the curve to find $y$ . $3(-1)^2 + y^4 + 6(-1) = 253$
Simplify Equation: Simplifying the equation, we get: $\newline$ $3(1) + y^4 - 6 = 253$ $\newline$ $3 + y^4 - 6 = 253$ $\newline$ $y^4 = 253 + 6 - 3$ $\newline$ $y^4 = 256$
Solve for $y$ : Taking the fourth root of both sides to solve for $y$ , we get: $\newline$ $y = \pm 4$ $\newline$ Since we are looking for the horizontal tangent line above the x-axis, we only consider the positive $y$ -value. $\newline$ $y = 4$
Equation of Tangent Line: We now have the point of tangency $(-1, 4)$ . The equation of a horizontal line is of the form $y = k$ , where $k$ is the $y$ -coordinate of the point through which the line passes. $\newline$ Therefore, the equation of the horizontal line tangent to the curve and above the $x$ -axis is $y = 4$ .

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