Question

graph the following curves and determine the location of any vertical tangent lines. a. x² + y² = 9 b. x² + y²+4x = 0 a. which of the following graphs displays the curve x² + y² = 9? a. b. c. d. determine the location of any vertical tangent lines. select the correct answer below. if necessary, fill in the answer box to complete your choice. a. the vertical tangent line(s) is(are) located at x = (type an integer or decimal. use a comma to separate answers as needed.) b. there are no vertical tangent lines.

Explanation:

Step1: Rewrite the equations in standard form

For the circle $x^{2}+y^{2}=9$, its center is at the origin $(0,0)$ and radius $r = 3$. For the circle $x^{2}+y^{2}+4x=0$, we complete the square: $(x + 2)^{2}+y^{2}-4=0$, or $(x + 2)^{2}+y^{2}=4$, with center $(-2,0)$ and radius $r = 2$.

Step2: Implicit - differentiate the equations

Differentiate $x^{2}+y^{2}=9$ with respect to $x$:
\[

$$\begin{align*} 2x + 2y\frac{dy}{dx}&=0\\ \frac{dy}{dx}&=-\frac{x}{y} \end{align*}$$

\]
A vertical tangent occurs when $\frac{dy}{dx}$ is undefined, i.e., $y = 0$. Substituting $y = 0$ into $x^{2}+y^{2}=9$, we get $x=\pm3$.
Differentiate $x^{2}+y^{2}+4x=0$ with respect to $x$:
\[

$$\begin{align*} 2x+2y\frac{dy}{dx}+4&=0\\ \frac{dy}{dx}&=-\frac{x + 2}{y} \end{align*}$$

\]
A vertical tangent occurs when $y = 0$. Substituting $y = 0$ into $x^{2}+y^{2}+4x=0$, we have $x^{2}+4x=0$, which factors to $x(x + 4)=0$. So $x=0$ or $x=-4$.

Answer:

For $x^{2}+y^{2}=9$, the vertical tangents are at $x = 3,x=-3$. For $x^{2}+y^{2}+4x=0$, the vertical tangents are at $x = 0,x=-4$.