Advanced Mathematical Concepts: Precalculus with Applications, Student Edition
Advanced Mathematical Concepts: Precalculus with Applications, Student Edition
1st Edition
ISBN: 9780078682278
Author: McGraw-Hill, Berchie Holliday
Publisher: Glencoe/McGraw-Hill
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Chapter 10.7, Problem 33E
To determine

Tofind:the graph of the equation and the graph of equation is degenerated case.

Expert Solution & Answer
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Answer to Problem 33E

The new equation so form is y=2x3 .

Explanation of Solution

Given:

  (x2)2(x+3)2=5(y+2) .

Concept used:

The general equation for conic section is:

  Ax2+Bxy+Cy2+Dx+Ey+F=0.

Where A,B,C,D,E and F are constant.

As by changing the values of some of the constants, the shape of the corresponding conic also changes.

If B24AC , is less than zero, if a conic exists, it will be either a circle or an ellipse.

If B24AC ,equals to zero, if a conic exists, it will be parabola.

If B24AC id greater than zero, if a conic exists, it will be hyperbola.

Calculation:

Consider the equation of the conic:

  (x2)2(x+3)2=5(y+2) .

Rearrange the equation in the general form of conic:

  Ax2+Bxy+Cy2+Dx+Ey+F=0.

  B=0 if conic parallel to co-ordinate axis else B0 .

  (x22.2.x+4)(x2+2.3.x+9)=5(y+2).x24x+4x26x9=5(y+2).10x5=5(y+2).2xy3=0.

Comparing the equation with general form of conic above:

  A=0,B=0,C=0,D=2,E=1,F=3.

Now,

  B24AC=04.0.0=0.

As in the given equation (x2)2(x+3)2=5(y+2) value of B24AC=0 so, at first glance the given equation may appear to be that of a parabola but a closer inspection reveals that this is a degenerate case.

So, by solving the equation (x2)2(x+3)2=5(y+2) to get:

  2xy3=0.y=2x3.

Now find the certain points to graph the straight line y=2x3 :

The graph of equation y=2x3 is:

  Advanced Mathematical Concepts: Precalculus with Applications, Student Edition, Chapter 10.7, Problem 33E

Hence, the new equation so form is y=2x3 .

Chapter 10 Solutions

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

Ch. 10.1 - Prob. 11CFUCh. 10.1 - Prob. 12ECh. 10.1 - Prob. 13ECh. 10.1 - Prob. 14ECh. 10.1 - Prob. 15ECh. 10.1 - Prob. 16ECh. 10.1 - Prob. 17ECh. 10.1 - Prob. 18ECh. 10.1 - Prob. 19ECh. 10.1 - Prob. 20ECh. 10.1 - Prob. 21ECh. 10.1 - Prob. 22ECh. 10.1 - Prob. 23ECh. 10.1 - Prob. 24ECh. 10.1 - Prob. 25ECh. 10.1 - Prob. 26ECh. 10.1 - Prob. 27ECh. 10.1 - Prob. 28ECh. 10.1 - Prob. 29ECh. 10.1 - Prob. 30ECh. 10.1 - Prob. 31ECh. 10.1 - Prob. 32ECh. 10.1 - Prob. 33ECh. 10.1 - Prob. 34ECh. 10.1 - Prob. 35ECh. 10.1 - Prob. 36ECh. 10.1 - Prob. 37ECh. 10.1 - Prob. 38ECh. 10.1 - Prob. 39ECh. 10.1 - Prob. 40ECh. 10.1 - Prob. 41ECh. 10.1 - Prob. 42ECh. 10.1 - Prob. 43ECh. 10.1 - Prob. 44ECh. 10.2 - Prob. 1CFUCh. 10.2 - Prob. 2CFUCh. 10.2 - Prob. 3CFUCh. 10.2 - Prob. 4CFUCh. 10.2 - Prob. 5CFUCh. 10.2 - Prob. 6CFUCh. 10.2 - Prob. 7CFUCh. 10.2 - Prob. 8CFUCh. 10.2 - Prob. 9CFUCh. 10.2 - Prob. 10CFUCh. 10.2 - Prob. 11CFUCh. 10.2 - Prob. 12CFUCh. 10.2 - Prob. 13CFUCh. 10.2 - Prob. 14CFUCh. 10.2 - Prob. 15ECh. 10.2 - 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