The function f(x)=x^3/x^2+ax+a has one vertical asymptote.what is a?
You need to remember that the form of equation of vertical asymptote is x = a and f(a) does not exist.
Hence, since the equation of the function is a fraction, then the function does not exist for the zeroes of denominator.
The problem provides the information that the function has one vertical asymptote, hence, the denominator has two equal roots.
You need to remember that the quadratic `ax^2 + bx + c = 0` has two equal roots if `Delta` is zero such that:
`Delta = b^2 - 4ac`
You need to identify a,b,c such that:
`a=1, b=a, c=a`
`Delta = a^2 - 4a`
You need to solve for a the equation `a^2 - 4a = 0` such that:
`a^2 - 4a = 0`
You need to factor out a such that:
`a(a - 4) = 0`
`a_1 = 0 `
`a_2 = 4`
Hence, evaluating the values of a for the function to have one vertical asymptote yields `a_1=0 ; a_2=4.`