Find the vertical asymtone: f(x)=`(x-2)/(2x^(2)+3x-2)`
To determine the value(s) of x for which the denominator is 0, solve `2x^2+3x-2=0`. Use the quadratic formula:
`x=(-b+-sqrt(b^2-4ac))/(2a)` for an equation of the form `ax^2+bx+c=0`
Substitute 3 for b, 2 for a and -2 for c.
The function is undefined for x=-2 and x=0.5, thus x=-2 and x=0.5 are the asymptotes.
Rational functions (functions with a ratio of polynomials) often have vertical asymptotes when the denominator equals zero.
In this case, we can factor the denominator:
`2x^2 + 3x - 2 = (2x-1)(x+2)`
Therefore the denominator equals zero AND we have vertical asymptotes when
`x = 1/2`