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Eleanor tried to find all the equations of vertical lines tangent to the curve given by 3x^(2)+24 x-y^(2)+2y=-44". " This is her solution: Step 1: Finding an expression for (dy)/(dx). (dy)/(dx)=(y-1)/(3(x+4)) Step 2: Forming a system of equations. {[3x^(2)+24 x-y^(2)+2y=-44],[3(x+4)=0],[y-1!=0]:} Step 3: Solving the system. There are no solutions to the system of equations, so there are no vertical lines tangent to the graph. Is Eleanor's solution correct? If not, at which step did she make a mistake? Choose 1 answer: (A) The solution is correct. (B) Step 1 is incorrect. (C) Step 2 is incorrect. (D) Step 3 is incorrect.

Eleanor tried to find all the equations of vertical lines tangent to the curve given by\newline3x2+24x−y2+2y=−44.  3 x^{2}+24 x-y^{2}+2 y=-44 \text {. } \newlineThis is her solution:\newlineStep 11: Finding an expression for dydx \frac{d y}{d x} .\newlinedydx=y−13(x+4) \frac{d y}{d x}=\frac{y-1}{3(x+4)} \newlineStep 22: Forming a system of equations.\newline{3x2+24x−y2+2y=−443(x+4)=0y−1≠0 \left\{\begin{array}{l} 3 x^{2}+24 x-y^{2}+2 y=-44 \\ 3(x+4)=0 \\ y-1 \neq 0 \end{array}\right. \newlineStep 33: Solving the system.\newlineThere are no solutions to the system of equations, so there are no vertical lines tangent to the graph.\newlineIs Eleanor's solution correct? If not, at which step did she make a mistake?\newlineChoose 11 answer:\newline(A) The solution is correct.\newline(B) Step 11 is incorrect.\newline(C) Step 2 \mathbf{2} is incorrect.\newline(D) Step 33 is incorrect.

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Q. Eleanor tried to find all the equations of vertical lines tangent to the curve given by\newline3x2+24x−y2+2y=−44.  3 x^{2}+24 x-y^{2}+2 y=-44 \text {. } \newlineThis is her solution:\newlineStep 11: Finding an expression for dydx \frac{d y}{d x} .\newlinedydx=y−13(x+4) \frac{d y}{d x}=\frac{y-1}{3(x+4)} \newlineStep 22: Forming a system of equations.\newline{3x2+24x−y2+2y=−443(x+4)=0y−1≠0 \left\{\begin{array}{l} 3 x^{2}+24 x-y^{2}+2 y=-44 \\ 3(x+4)=0 \\ y-1 \neq 0 \end{array}\right. \newlineStep 33: Solving the system.\newlineThere are no solutions to the system of equations, so there are no vertical lines tangent to the graph.\newlineIs Eleanor's solution correct? If not, at which step did she make a mistake?\newlineChoose 11 answer:\newline(A) The solution is correct.\newline(B) Step 11 is incorrect.\newline(C) Step 2 \mathbf{2} is incorrect.\newline(D) Step 33 is incorrect.
  1. Differentiate and Solve: To find the slope of the tangent lines to the curve, we need to differentiate the given equation implicitly with respect to xx to find dydx\frac{dy}{dx}. Differentiating each term with respect to xx, we get: ddx(3x2)+ddx(24x)−ddx(y2)+ddx(2y)=ddx(−44)\frac{d}{dx}(3x^2) + \frac{d}{dx}(24x) - \frac{d}{dx}(y^2) + \frac{d}{dx}(2y) = \frac{d}{dx}(-44) 6x+24−2ydydx+2dydx=06x + 24 - 2y\frac{dy}{dx} + 2\frac{dy}{dx} = 0 Now, we solve for dydx\frac{dy}{dx}: dydx(2−2y)=−6x−24\frac{dy}{dx}(2 - 2y) = -6x - 24 dydx=−6x−242−2y\frac{dy}{dx} = \frac{-6x - 24}{2 - 2y}
  2. Find Vertical Tangent Condition: Vertical lines have an undefined slope, which means that dydx\frac{dy}{dx} would be undefined. This happens when the denominator of the expression for dydx\frac{dy}{dx} is zero. Therefore, we set the denominator equal to zero to find the xx-values where the slope is undefined:\newline2−2y=02 - 2y = 0\newliney=1y = 1\newlineNow we form a system of equations with the original equation and the condition for a vertical tangent y=1y = 1:\newline{3x2+24x−y2+2y=−44 y=1\begin{cases} 3x^2 + 24x - y^2 + 2y = -44 \ y = 1 \end{cases}
  3. Substitute and Solve: We substitute y=1y = 1 into the original equation to find the xx-values where the vertical tangents occur:\newline3x2+24x−(1)2+2(1)=−443x^2 + 24x - (1)^2 + 2(1) = -44\newline3x2+24x−1+2=−443x^2 + 24x - 1 + 2 = -44\newline3x2+24x+1=−443x^2 + 24x + 1 = -44\newline3x2+24x+45=03x^2 + 24x + 45 = 0\newlineNow we solve for xx:\newlinex2+8x+15=0x^2 + 8x + 15 = 0\newline(x+5)(x+3)=0(x + 5)(x + 3) = 0\newlinex=−5x = -5 or x=−3x = -3\newlineThese are the xx-values where the vertical tangents occur. The equations of the vertical lines are x=−5x = -5 and x=−3x = -3.

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