Question

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:

1. First, recall the condition for a vertical tangent line:

• For a curve given by an implicit - function \(F(x,y)=0\), the slope of the tangent line is \(\frac{dy}{dx}\). A vertical tangent line occurs when \(\frac{dy}{dx}\) is undefined, which is equivalent to the denominator of \(\frac{dy}{dx}\) being zero (assuming \(\frac{dy}{dx}\) is expressed as a rational function).
• For a circle equation \((x - a)^2+(y - b)^2=r^2\), we can use implicit differentiation. Let's assume the general case. If we have the equation \(x^{2}+y^{2}+5x = 0\).
• Differentiate both sides of the equation with respect to \(x\):
• Using the sum - rule and the chain - rule, \(\frac{d}{dx}(x^{2})+\frac{d}{dx}(y^{2})+\frac{d}{dx}(5x)=\frac{d}{dx}(0)\).
• We know that \(\frac{d}{dx}(x^{2}) = 2x\), \(\frac{d}{dx}(y^{2})=2y\frac{dy}{dx}\) (by the chain - rule, since \(y\) is a function of \(x\)), and \(\frac{d}{dx}(5x)=5\), \(\frac{d}{dx}(0)=0\).
• So, \(2x + 2y\frac{dy}{dx}+5 = 0\).
• Solve for \(\frac{dy}{dx}\):
• First, isolate the terms with \(\frac{dy}{dx}\): \(2y\frac{dy}{dx}=-2x - 5\).
• Then, \(\frac{dy}{dx}=\frac{-2x - 5}{2y}\).

1. Then, find when \(\frac{dy}{dx}\) is undefined:

• \(\frac{dy}{dx}\) is undefined when \(y = 0\).
• Substitute \(y = 0\) into the original equation \(x^{2}+y^{2}+5x = 0\).
• When \(y = 0\), the equation becomes \(x^{2}+5x=0\).
• Factor out an \(x\): \(x(x + 5)=0\).
• Set each factor equal to zero:
• If \(x=0\), then \(x(x + 5)=0\); if \(x=-5\), then \(x(x + 5)=0\).

So the vertical tangent line(s) is(are) located at \(x=-5,0\).

Step1: Implicit differentiation

Differentiate \(x^{2}+y^{2}+5x = 0\) with respect to \(x\) to get \(2x + 2y\frac{dy}{dx}+5 = 0\).

Step2: Solve for \(\frac{dy}{dx}\)

Isolate \(\frac{dy}{dx}\) to obtain \(\frac{dy}{dx}=\frac{-2x - 5}{2y}\).

Step3: Find when \(\frac{dy}{dx}\) is undefined

Set \(y = 0\) and substitute into the original equation \(x^{2}+y^{2}+5x = 0\), then factor \(x^{2}+5x=0\) to \(x(x + 5)=0\) and solve for \(x\).

Answer:

\(x=-5,0\)