(x^2+x-2) / (x^2-3x-4)
X and y intercepts?
Horizontal Asymptotes?
Vertical Asymptotes?
y = (x-1)(x+2) / [ (x-4)(x+1)]
when x = +1 , y = 0
when x = -2 , y = 0
when x = 0 , y = -2/-4 = 1/2
vertical when x = 4 or x = -1
= x^2/x^2 --> 1 when x is big positive or big negative
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(x^2+x-2) / (x^2-3x-4)
here:
http://www.mathsisfun.com/data/function-grapher.php
To find the x-intercepts, we need to solve the equation f(x) = 0. For x-intercepts, the numerator (x^2 + x - 2) equals zero since y = 0 on the x-axis.
So, let's solve for x:
x^2 + x - 2 = 0
We can factor this quadratic equation:
(x - 1)(x + 2) = 0
Setting each factor to zero gives us two solutions:
x - 1 = 0 => x = 1
x + 2 = 0 => x = -2
Therefore, the x-intercepts are (1, 0) and (-2, 0).
To find the y-intercept, we need to find the value of f(x) when x = 0. So, substitute x = 0 into the equation:
f(0) = (0^2 + 0 - 2) / (0^2 - 3(0) - 4)
= -2 / -4
= 1/2
Therefore, the y-intercept is (0, 1/2).
Moving on to the horizontal asymptotes, we need to examine the behavior of the function as x approaches positive or negative infinity. To determine the horizontal asymptotes, we can compare the degrees of the numerator and denominator polynomials.
The degree of the numerator is 2 (since the highest power term is x^2) and the degree of the denominator is also 2 (since the highest power term is x^2). When the degrees are the same, we compare the coefficients of the highest power terms.
For this equation, the coefficient of x^2 in both the numerator and denominator is 1. Therefore, the horizontal asymptote is y = 1.
Finally, let's find the vertical asymptotes. Vertical asymptotes occur when the denominator of the fraction equals zero. To find these points, we set the denominator (x^2 - 3x - 4) equal to zero and solve for x:
x^2 - 3x - 4 = 0
This quadratic equation cannot be factored, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -3, and c = -4. Plugging in these values, we get:
x = (-(-3) ± √((-3)^2 - 4(1)(-4))) / (2 * 1)
x = (3 ± √(9 + 16)) / 2
x = (3 ± √25) / 2
x = (3 ± 5) / 2
This gives us two possible solutions:
x = (3 + 5) / 2 => x = 8/2 => x = 4
x = (3 - 5) / 2 => x = -2/2 => x = -1
Therefore, the vertical asymptotes are x = 4 and x = -1.