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Chapter 2: Problem 64

Find the points at which the graph of the equation has a vertical or horizontal tangent line. $$ 4 x^{2}+y^{2}-8 x+4 y+4=0 $$

Short Answer

Expert verified
The given graph has a vertical tangent at the point (1, -2) and a horizontal tangent at the point (2, -2).

Step by step solution

01

Write the equation in standard form

The given equation is \(4x^2+y^2-8x+4y+4\) = 0. In order to simplify, it needs to be written in the standard form. Completing the square for both x and y variables, we obtain: \((2x-2)^2+(y+2)^2 = 0\).
02

Differentiate the equation

Next, you need to find the derivative of the function. Implicit differentiation is applied, considering y as a function of x, keeping in mind the chain rule. The derivative of \((2x-2)^2+(y+2)^2\) with respect to x is: \( 4(2x - 2) + 2(y + 2)y' = 0\). You can simplify this to \( 4x + y'y = 0 \). Then, we calculate y' = -4x/y.
03

Find the horizontal and vertical tangents

The slope of the tangent line is given by the derivative. When the derivative equals 0, the tangent line is horizontal. And, when the derivative is undefined, the tangent line is vertical. Setting y'= 0 and solving for y, we result in y = 0. And if y != 0, x = 0. Substitute x = 0 and y = 0 into the original equation, the points where the graph has horizontal and vertical tangents can be found.
04

Conclusion

After substituting these values, the result is: the graph has a vertical tangent at the point (1, -2) and a horizontal tangent at the point (2, -2).

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Most popular questions from this chapter

In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{y=2 \tan ^{3} x} \quad \frac{\text { Point }}{\left(\frac{\pi}{4}, 2\right)}\)

In Exercises \(75-80\), evaluate the derivative of the function at the indicated point. Use a graphing utility to verify your result. \(\frac{\text { Function }}{y=\frac{1}{x}+\sqrt{\cos x}} \quad \frac{\text { Point }}{\left(\frac{\pi}{2}, \frac{2}{\pi}\right)}\)

In Exercises 37 and 38 , the derivative of the function has the same sign for all \(x\) in its domain, but the function is not one-to-one. Explain. $$ f(x)=\frac{x}{x^{2}-4} $$

Let \(u\) be a differentiable function of \(x\). Use the fact that \(|u|=\sqrt{u^{2}}\) to prove that \(\frac{d}{d x}[|u|]=u^{\prime} \frac{u}{|u|}, \quad u \neq 0\).

Evaluate the second derivative of the function at the given point. Use a computer algebra system to verify your result. \(f(x)=\frac{1}{\sqrt{x+4}}, \quad\left(0, \frac{1}{2}\right)\)
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