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$Kendall tried to find all the equations of vertical lines tangent to the curve given by x^(2)+2xy^(2)=25. This is her solution: Step 1: Finding an expression for (dy)/(dx). (dy)/(dx)=(-x-y^(2))/(2xy) Step 2: Forming a system of equations. {[x^(2)+2xy^(2)=25],[2xy=0],[-x-y^(2)!=0]:} Step 3: Solving the system. x=-5,x=0, and x=5 Is Kendall's solution correct? If not, at which step did she make a mistake? Choose 1 answer: (A) The solution is correct. (B) Step 1 is incorrect. (C) Step 2 is incorrect. (D) Step 3 is incorrect.$

Kendall tried to find all the equations of vertical lines tangent to the curve given by $x^{2}+2 x y^{2}=25$ . This is her solution: $\newline$ Step $1$ : Finding an expression for $\frac{d y}{d x}$ . $\newline$ $\frac{d y}{d x}=\frac{-x-y^{2}}{2 x y}$ $\newline$ Step $2$ : Forming a system of equations. $\newline$ $\left\{\begin{array}{l} x^{2}+2 x y^{2}=25 \\ 2 x y=0 \\ -x-y^{2} \neq 0 \end{array}\right.$ $\newline$ Step $3$ : Solving the system. $\newline$ $x=-5, x=0$ , and $x=5$ $\newline$ Is Kendall's solution correct? If not, at which step did she make a mistake? $\newline$ Choose $1$ answer: $\newline$ (A) The solution is correct. $\newline$ (B) Step $1$ is incorrect. $\newline$ (C) Step $2$ is incorrect. $\newline$ (D) Step $3$ is incorrect.

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Algebra 1

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Q. Kendall tried to find all the equations of vertical lines tangent to the curve given by

x^{2}+2 x y^{2}=25

. This is her solution:

\newline

Step

1

: Finding an expression for

\frac{d y}{d x}

\newline

\frac{d y}{d x}=\frac{-x-y^{2}}{2 x y}

\newline

Step

2

: Forming a system of equations.

\newline

\left\{\begin{array}{l} x^{2}+2 x y^{2}=25 \\ 2 x y=0 \\ -x-y^{2} \neq 0 \end{array}\right.

\newline

Step

3

: Solving the system.

\newline

x=-5, x=0

, and

x=5

\newline

Is Kendall's solution correct? If not, at which step did she make a mistake?

\newline

Choose

1

answer:

\newline

(A) The solution is correct.

\newline

(B) Step

1

is incorrect.

\newline

2

is incorrect.

\newline

(D) Step

3

is incorrect.

Forming Equations: Forming a system of equations. $\newline$ Kendall's system of equations is based on the conditions for a vertical tangent line. For a vertical line, the slope is undefined, which means $\frac{dy}{dx}$ should be undefined. This occurs when the denominator of our derivative is zero, so $2xy = 0$ . Additionally, the numerator must not be zero to avoid the indeterminate form $\frac{0}{0}$ , so $-x - y^2 \neq 0$ . $\newline$ Kendall's system of equations is: $\newline$ $1$ . $x^2 + 2xy^2 = 25$ (original equation) $\newline$ $2$ . $2xy = 0$ (condition for vertical tangent) $\newline$ $3$ . $-x - y^2 \neq 0$ (to avoid indeterminate form) $\newline$ Kendall's system of equations is correct.
Solving the System: Solving the system. $\newline$ Kendall's solution to the system is $x = -5$ , $x = 0$ , and $x = 5$ . $\newline$ Let's check these solutions: $\newline$ From the second equation, $2xy = 0$ , we can see that either $x = 0$ or $y = 0$ for the equation to hold true. However, if $y = 0$ , then the first equation becomes $x^2 = 25$ , which gives us $x = \pm5$ . If $x = 0$ , then the first equation becomes $x = 0$ $0$ , which is not true. Therefore, $x = 0$ $1$ cannot be $x = 0$ $2$ . $\newline$ Kendall's inclusion of $x = 0$ as a solution is incorrect. The correct solutions should only be $x = -5$ and $x = 5$ , as these are the x-values where the curve $x = 0$ $6$ has vertical tangents. $\newline$ Kendall made a mistake in this step.