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Vertical and Horizontal Tangent Lines In Exercises 61 and \(62,\) 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 points at which the graph of the equation \(4 x^{2}+y^{2}-8 x+4 y+4=0\) has a vertical tangent line are (0,2), (0,-2), (2,2) and (2,-2) and the points at which it has a horizontal tangent line are (1,2) and (1,-2).

Step by step solution

01

Rewrite the equation

Rearrange the given equation \(4x^{2}+y^{2}-8x+4y+4=0\) as a function of x. This results in \(y(x)=\pm \sqrt{8x-4x^{2}-4}\).
02

Find derivative of the function

Differentiate the equation obtained from step 1 with respect to x, \(y'(x)=\pm \frac{8-8x}{2\sqrt{8x-4x^{2}-4}}\).
03

Find x where y'(x)=0

The solutions to the equation \(y'(x)=0\) are the x-values of the points with a horizontal tangent. To find this, set the numerator of y'(x) equal to zero, \(8-8x=0\), which gives \(x=1\). Substitute this value into \(y(x)\) to find the corresponding y-values. This gives \(y=\pm \sqrt{8-4}= \pm 2\). So, the points are (1,2) and (1,-2)
04

Find x where y'(x) is undefined

The solutions to the equation \(y'(x)\) is undefined are the x-values of the points with a vertical tangent. To find this, set the denominator of y'(x) equal to zero, \(8x-4x^{2}-4=0\). This is a quadratic equation that can be solved by factoring, quadratic formula or completing the square. The solutions are \(x=0\) and \(x=2\). Substitute these values in the original equation to find the corresponding y-values. This gives (0,2) and (0,-2), (2,2) and (2,-2).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Lines
Tangent lines are crucial in understanding the slope and direction of a curve at a given point. Imagine them as the lines that lightly touch the curve at one point without cutting across it. For a visual perspective, think of riding on a bicycle path that follows a curvy line; the tangent line at any point on this path would be like a straight, narrow road that just kisses the curve. Specialty in tangent lines arises when they are horizontal or vertical, which informs us many valuable insights about the behavior of the graph at those points. When the tangent line is horizontal, it means the slope at that point is zero, indicating that the curve is flat, like the top of a hill. On the other hand, a vertical tangent suggests an undefined slope, reflecting a steep rise or fall at that point.
Derivative
The derivative is a key concept in calculus. Simply put, it describes the rate at which a function changes. Imagine a derivative as a speedometer for functions, indicating how fast or slow the 'car' (function) is moving at any point. In mathematical terms, it's the slope of the curve at a specific point—the steeper the slope, the faster the change. Calculating derivatives lets you find tangent lines to graphs, unlocking secrets about the graph's behavior at specific points. If we have a function y = f(x), the derivative, denoted as f'(x) or \( \frac{dy}{dx} \), helps determine the function's increasing or decreasing nature, enabling us to find points where tangent lines are horizontal or vertical.
Quadratic Equations
Quadratic equations are vital in solving various mathematical problems, set in the form \( ax^2 + bx + c = 0 \). These equations form a parabola when graphed on a coordinate plane. They are characterized by a gentle, symmetrical curve that opens either up or down depending on the coefficient of \( x^2 \). Solving quadratic equations can be done using several methods: factoring, using the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \), or completing the square. These solutions give the x-value(s) where the curve crosses the x-axis. In our exercise's context, solving the quadratic equation helps find x-values that lead to either undefined or horizontal tangent lines, revealing important graph features and behaviors.
Differentiation
Differentiation is the process of finding a derivative, helping to understand how a function’s output changes with respect to its input. It breaks down complex motions and changes into clear, simpler parts, crafting a picture of how fast functions are climbing up or sliding down. Imagine watching a movie and differentiation as a tool that lets you analyze each frame to understand the action. It's this method that tells us the slope of the tangent and where they're horizontal or vertical. Through differentiating a given function, you can explore the exact points of interest on a curve where these changes are most significant. Differentiation not only simplifies the diving into calculus problems but is also a gateway to advanced mathematical applications in real life, ranging from physics to economics.

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Most popular questions from this chapter

The number \(N\) of bacteria in a culure after t days is modeled by \(N=400\left[1-\frac{3}{\left(t^{2}+2\right)^{2}}\right]\) Find the rate of change of \(N\) with respect to \(t\) when (a) \(t=0,\) (b) \(t=1\) (c) \(t=2,\) (d) \(t=3\)and (e) \(t=4 .\) (f) What can you conclude?

FOR FURTHER INFORMATION For more information on the mathematics of moving ladders, see the article "The Falling Ladder Paradox" by Paul Scholten and Andrew Simoson in The College Mathematics Journal. To view this article, go to MathArticles.com. Construction A construction worker pulls a five-meter plank up the side of a building under construction by means of a rope tied to one end of the plank (see figure). Assume the opposite end of the plank follows a path perpendicular to the wall of the building and the worker pulls the rope at a rate of 0.15 meter per second. How fast is the end of the plank sliding along the ground when it is 2.5 meters from the wall of the building?

Pendulum A 15 -centimeter pendulum moves according to the equation \(\theta=0.2 \cos 8 t,\) where \(\theta\) is the angular displacement from the vertical in radians and \(t\) is the time in seconds. Determine the maximum angular displacement and the rate of change of \(\theta\) when \(t=3\) seconds.

Finding the Slope of a Graph In Exercises \(63-70\) , find the slope of the graph of the function at the given point. Use the derivative feature of a graphing utility to confirm your results. $$y=\frac{4}{(x+2)^{2}}, \quad(0,1)$$

Horizontal Tangent Line In Exercises \(73-76\) determine the point(s) at which the graph of the function has a horizontal tangent line. \(f(x)=\frac{x-4}{x^{2}-7}\)

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