Question

Identify the horizontal asymptote of f(x) = quantity 2 x minus 1 over quantity x squared minus 7 x plus 3. y = 0 y = 1 over 2 y = 2 no horizontal asymptote

Answer 1

The horizontal asymptote is the value at the y-axis where the graph approaches the line but not necessarily touching it. Hence, the asymptotic characteristic of the graph. The standard form of a function in fraction form is y = (ax^n +...)/(bx^m+...). There are rules to follow to determine the horizontal asymptote of a function.
1) if n = m , then the horizontal equation is y = a/b
2) if n>m, then there is no horizontal equation
3) if n<m, then the horizontal equation is the x axis ; y = 0.

The function given falls on the third rule hence the horizontal asymptote of the function is at y = 0.

Answer 2

Answer with explanation:

The function is

$f(x)=\frac{2 x-1}{x^2-7 x-3}$

→Horizontal asymptote can be calculated as:

$1.y= \lim_{x \to 0} \frac{2 x-1}{x^2-7 x-3}\\\\2.y=\lim_{x \to 0}\frac{2 -\frac{1}{x}}{x-7-\frac{3}{x}}\\\\3. y=\frac{2}{\text{-Infinity}}\\\\4.y=0$

In Second step dividing numerator and denominator by ,x.

→y=0, is the  horizontal asymptote.