Write an equation that would have a hole at x = 4 and a vertical asympote at x = -2.
My answer was f(x)= (x-4)/(x+2)
If it is not correct can someone please help me solve it and explain it to me? Thanks!
a hole is a disruption at a single point
... try [(x - 4) / (x - 4)]
... the division by zero causes the discontinuity, but the equal numerator means no other point is affected
f(x) = (x - 4) / [(x - 4) (x + 2)]
Ohhh I see thanks Scott!
Sorry I actually have one more question. After solving for your equation the vertical asymptote would actually be 2 and not -2?? So can you still help me create the correct equation
Would it be f(x) = (x-4) / (x-4)(x+2)???
Scott was right, the denominator causing the vertical asymptote at x = -2 is
x+2
final answer:
f(x) = (x-4)/( (x-4)(x+2) )
Your answer is correct! The equation f(x) = (x-4)/(x+2) does indeed have a hole at x = 4 and a vertical asymptote at x = -2.
To understand why this equation has these features, let's break it down:
1. Hole at x = 4:
In order to have a hole at x = 4, there must be a factor of (x-4) in both the numerator and the denominator. This ensures that the function is undefined at x = 4, creating a hole in the graph. Therefore, the numerator (x-4) provides this requirement.
2. Vertical asymptote at x = -2:
A vertical asymptote occurs when the denominator of the rational function equals zero. In this case, the denominator (x+2) becomes zero when x = -2. So, the vertical asymptote is located at x = -2.
Combining these factors, the equation f(x) = (x-4)/(x+2) satisfies both conditions: a hole at x = 4 and a vertical asymptote at x = -2.