A Function of x

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 2, Section 2.1

Determine whether the equation defines y as a function of x.

See attachment.

[Math Tutor]

35. [tex]y^2 = 4 - x^2[/tex]

Solve for y:

[tex]y = \pm\sqrt{4 - x^2}[/tex]

The [tex]\pm[/tex] is the problem. Take [tex]x = 0[/tex]: then [tex]y^2 = 4[/tex], so [tex]y = 2[/tex] and [tex]y = -2[/tex]. One value of x gives two different values of y, so the vertical line test fails.

Therefore the equation does NOT define y as a function of x. (Its graph is the circle [tex]x^2 + y^2 = 4[/tex], a circle of radius 2 centered at the origin.)

[nycmath]

35. [tex]y^2 = 4 - x^2[/tex]

Solve for y:

[tex]y = \pm\sqrt{4 - x^2}[/tex]

The [tex]\pm[/tex] is the problem. Take [tex]x = 0[/tex]: then [tex]y^2 = 4[/tex], so [tex]y = 2[/tex] and [tex]y = -2[/tex]. One value of x gives two different values of y, so the vertical line test fails.

Therefore the equation does NOT define y as a function of x. (Its graph is the circle [tex]x^2 + y^2 = 4[/tex], a circle of radius 2 centered at the origin.)

I just saw this reply. Thank you. You are the best.