Consider the curve given by the equation 3x^(2)+6x-y^(2)+8y=16". It can " be shown that (dy)/(dx)=(3(x+1))/(y-4). Find the y-coordinate of the point where the line tangent to the curve is the y-axis. y=

Find Tangent Line: To find the y-coordinate of the point where the tangent line is the y-axis, we need to find the point where the slope of the tangent (dy/dx) is undefined, which occurs when the denominator of the derivative is zero.Set Derivative Equal to Zero: Set the denominator of the derivative (dy/dx) = (3(x+1))/(y-4) equal to zero to find the x-coordinate of the point where the tangent line is the y-axis. y - 4 = 0 y = 4Determine Coordinates: Now that we have the y-coordinate, we need to find the corresponding x-coordinate. Since the tangent line is the y-axis, the x-coordinate must be 0.Verify Point on Curve: Substitute x = 0 and y = 4 into the original equation to check if this point lies on the curve. 3(0)^2 + 6(0) - (4)^2 + 8(4) = 16 0 + 0 - 16 + 32 = 16 16 = 16 This confirms that the point (0, 4) is on the curve.

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Consider the curve given by the equation

3x^(2)+6x-y^(2)+8y=16". It can "
be shown that
(dy)/(dx)=(3(x+1))/(y-4).
Find the
y-coordinate of the point where the line tangent to the curve is the
y-axis.

y=

Consider the curve given by the equation $3 x^{2}+6 x-y^{2}+8 y=16.$ It can be shown that $\frac{d y}{d x}=\frac{3(x+1)}{y-4}$ . $\newline$ Find the $y$ -coordinate of the point where the line tangent to the curve is the $y$ -axis. $\newline$ $y=$

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Algebra 1

Transformations of quadratic functions

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Q. Consider the curve given by the equation

3 x^{2}+6 x-y^{2}+8 y=16.

It can be shown that

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

\newline

Find the

y

-coordinate of the point where the line tangent to the curve is the

y

-axis.

\newline

y=

Find Tangent Line: To find the $y$ -coordinate of the point where the tangent line is the $y$ -axis, we need to find the point where the slope of the tangent ( $\frac{dy}{dx}$ ) is undefined, which occurs when the denominator of the derivative is zero.
Set Derivative Equal to Zero: Set the denominator of the derivative $\frac{dy}{dx} = \frac{3(x+1)}{y-4}$ equal to zero to find the $x$ -coordinate of the point where the tangent line is the y-axis. $\newline$ $y - 4 = 0$ $\newline$ $y = 4$
Determine Coordinates: Now that we have the $y$ -coordinate, we need to find the corresponding $x$ -coordinate. Since the tangent line is the $y$ -axis, the $x$ -coordinate must be $0$ .
Verify Point on Curve: Substitute $x = 0$ and $y = 4$ into the original equation to check if this point lies on the curve. $3(0)^2 + 6(0) - (4)^2 + 8(4) = 16$ $0 + 0 - 16 + 32 = 16$ $16 = 16$ This confirms that the point $(0, 4)$ is on the curve.