AP Calculus AB $\newline$ AP Exam Review Free Response $6$ $\newline$ This Question is ${ }^{* *}$ CALCULATOR INACTIVE** $\newline$ Please show all work on page $2 \& 3$ $\newline$ The equation of the implicitly defined Tschirnhausen's Cubic is $y^{2}-3 x^{2}-x^{3}=0$ . The graph of Tschirnhausen's Cubic is shown below. $\newline$ (a) Find $\frac{d y}{d x}$ . Show all work! $\newline$ (b) Find the equation of the tangent line to Tschirnhausen's Cubic at the point $(1,-2)$ and use it to approximate the $y$ value of the curve where $x=1.1$ . $\newline$ (c) At what point(s) does this curve have a horizontal tangent? Give both the $x$ - and $y$ -coordinates. Show all work.

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Q. AP Calculus AB

\newline

AP Exam Review Free Response

6

\newline

This Question is

{ }^{* *}

CALCULATOR INACTIVE**

\newline

Please show all work on page

2 \& 3

\newline

The equation of the implicitly defined Tschirnhausen's Cubic is

y^{2}-3 x^{2}-x^{3}=0

. The graph of Tschirnhausen's Cubic is shown below.

\newline

(a) Find

\frac{d y}{d x}

. Show all work!

\newline

(b) Find the equation of the tangent line to Tschirnhausen's Cubic at the point

(1,-2)

and use it to approximate the

y

value of the curve where

x=1.1

\newline

x

- and

y

-coordinates. Show all work.

Implicit Differentiation: Differentiate the given equation implicitly to find $(\frac{dy}{dx})$ . Given equation: $y^2 - 3x^2 - x^3 = 0$ . Differentiate both sides with respect to $x$ : $2y(\frac{dy}{dx}) - 6x - 3x^2(\frac{dx}{dx}) = 0$ . Simplify and solve for $(\frac{dy}{dx})$ : $2y(\frac{dy}{dx}) - 6x - 3x^2 = 0$ , $(\frac{dy}{dx}) = \frac{6x + 3x^2}{2y}$ .
Slope Calculation: Substitute $x = 1$ and $y = -2$ into the derivative to find the slope of the tangent line at the point $(1, -2)$ . $(\frac{dy}{dx}) = \frac{(6(1) + 3(1)^2)}{(2(-2))},$ $(\frac{dy}{dx}) = \frac{(6 + 3)}{(-4)},$ $(\frac{dy}{dx}) = \frac{9}{-4},$ $(\frac{dy}{dx}) = -2.25.$
Tangent Line Equation: Use the point-slope form to find the equation of the tangent line at $(1, -2)$ .
Point-slope form: $y - y_1 = m(x - x_1)$ ,
$y + 2 = -2.25(x - 1)$ ,
$y + 2 = -2.25x + 2.25$ ,
$y = -2.25x + 0.25$ .
Y-Value Approximation: Substitute $x = 1.1$ into the tangent line equation to approximate the y-value. $\newline$ $y = -2.25(1.1) + 0.25$ , $\newline$ $y = -2.475 + 0.25$ , $\newline$ $y = -2.225$ .
Horizontal Tangent Points: Find points where the curve has a horizontal tangent, i.e., $(\frac{dy}{dx}) = 0$ . Set the derivative equal to zero: $\frac{(6x + 3x^2)}{(2y)} = 0$ , $6x + 3x^2 = 0$ , $x(6 + 3x) = 0$ , $x = 0$ or $x = -2$ . Substitute back into the original equation to find $y$ : For $x = 0$ , $y^2 - 3(0)^2 - 0^3 = 0$ , $y = 0$ . For $x = -2$ , $\frac{(6x + 3x^2)}{(2y)} = 0$ $1$ , $\frac{(6x + 3x^2)}{(2y)} = 0$ $2$ , $\frac{(6x + 3x^2)}{(2y)} = 0$ $3$ , $\frac{(6x + 3x^2)}{(2y)} = 0$ $4$ .