07-18-2006, 11:07 AM

I am a little rusty on this topic....How do you find x-intercept(s) of the equation

y=(x^3+3x^2)/ (2*(x^2+x-6))??

y=(x^3+3x^2)/ (2*(x^2+x-6))??

View Full Version : finding x-intercept(s) of y = (x^3 + 3x^2)/(2(x^2 + x - 6))

07-18-2006, 11:07 AM

I am a little rusty on this topic....How do you find x-intercept(s) of the equation

y=(x^3+3x^2)/ (2*(x^2+x-6))??

y=(x^3+3x^2)/ (2*(x^2+x-6))??

stapel

07-18-2006, 11:15 AM

Find the x-intercept(s) in the usual manner: Plug zero in for "y", and solve the resulting equation.

In this case, of course, it is helpful to remember that a fraction is zero when its numerator is zero. :wink:

Eliz.

In this case, of course, it is helpful to remember that a fraction is zero when its numerator is zero. :wink:

Eliz.

skeeter

07-18-2006, 11:21 AM

y = (x^3 + 3x^2)/ [2*(x^2 + x - 6)]

y = x^2(x + 3)/[2(x + 3)(x - 2)]

the (x+3) in the numerator and denominator cancel, creating a removable discontinuity (a hole) at x = -3. the numerator equals zero at x = 0, so a single x-intercept is at the origin, (0,0).

y = x^2(x + 3)/[2(x + 3)(x - 2)]

the (x+3) in the numerator and denominator cancel, creating a removable discontinuity (a hole) at x = -3. the numerator equals zero at x = 0, so a single x-intercept is at the origin, (0,0).

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