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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

User Wenny
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Answer with explanation:

The function is


f(x)=(2 x-1)/(x^2-7 x-3)

→Horizontal asymptote can be calculated as:


1.y= \lim_(x \to 0) (2 x-1)/(x^2-7 x-3)\\\\2.y=\lim_(x \to 0)(2 -(1)/(x))/(x-7-(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.

User DannyT
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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.
User Charu Maheshwari
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