BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1.9, Problem 27E
To determine

To graph: The circle x2+y2=9 by solve for y .

Expert Solution

Explanation of Solution

Given information:

The equation of circle x2+y2=9 .

Graph:

The graph of circle x2+y2=9 by solving for y .

Subtract x2 from the equation of circle,

  x2x2+y2=9x2y2=9x2

Take under root on both side of the equation.

  y=±9x2

From the two equations graph of circle is described:

  y=+9x2,y=9x2

The graph of the equation y=+9x2 represents upper half of the circle because y is always greater than or equal to 0.

  9x209x209x2±3x

So, the values of x is less than or equal to ±3 .

Thus, the graph appropriate in the [3,3]×[0,3] viewing rectangle

The graph of the equation can be sketched using the table,

    xy
    03
    25
    30
    18

The graph of equation is provided below,

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 1.9, Problem 27E , additional homework tip  1

The graph of the equation y=9x2 represents lower half of the circle because y is always less than or equal to 0.

  9x209x209x2±3x

So, the values of x is greater than or equal to ±3 .

Thus, the graph appropriate in the [3,3]×[-3,0] viewing rectangle

The graph of the equation y=9x2 can be sketched using the table,

    xy
    03
    25
    30
    18

The graph of equation is provided below,

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 1.9, Problem 27E , additional homework tip  2

The graph of both equations in same figure is provided below,

The graph appropriate in the [3,3]×[-3,3] viewing rectangle

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 1.9, Problem 27E , additional homework tip  3

Interpretation:

The graph of an equation in a viewing screen is a viewing rectangle.

The x -values to range from a minimum value of xmin =a to a maximum value of xmax =b

The y -values to range from minimum value of ymin =c to a maximum value of ymax =d

Then, the display portion of the graph lies in the rectangle [a,b]×[c,d]={(x,y)|axb,cyd}

The graph of y=+9x2 appropriate in the [3,3]×[0,3] viewing rectangle and the graph of this equation lies on the above part or the graph make a half circle.

The graph of y=9x2 appropriate in the [3,3]×[-3,0] viewing rectangle and the graph of this equation lies on the lower part or the graph make a half circle.

The graph of x2+y2=9 appropriate in the [3,3]×[-3,3] viewing rectangle and the graph make a circle.

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