Question

determine each feature of the graph of the given function.
$f(x) = \frac{4x - 20}{x^2 - 5x}$
answer attempt 1 out of 3
horizontal asymptote: $y = \square$ $\boxed{\text{no horizontal asymptote}}$
vertical asymptote: $x = \square$ $\boxed{\text{no vertical asymptote}}$
x-intercept: $(\square, 0)$ $\boxed{\text{no x-intercept}}$
y-intercept: $(0, \square)$ $\boxed{\text{no y-intercept}}$
hole: $(\square, \square)$ $\boxed{\text{no hole}}$

Explanation:

Step1: Simplify the function

First, factor the numerator and the denominator. The numerator \(4x - 20\) can be factored as \(4(x - 5)\), and the denominator \(x^2 - 5x\) can be factored as \(x(x - 5)\). So the function becomes:

$$ f(x)=\frac{4(x - 5)}{x(x - 5)} $$

We can cancel out the common factor \((x - 5)\) (for \(x
eq5\)), so the simplified function is \(f(x)=\frac{4}{x}\) with a hole at \(x = 5\) (since \(x = 5\) makes the original denominator zero but is canceled out).

Step2: Find Horizontal Asymptote

For a rational function \(f(x)=\frac{N(x)}{D(x)}\), if the degree of \(N(x)\) (degree of numerator) is less than the degree of \(D(x)\) (degree of denominator), the horizontal asymptote is \(y = 0\). In the simplified function \(f(x)=\frac{4}{x}\), the degree of numerator (0, since \(4 = 4x^0\)) is less than the degree of denominator (1, since \(x=x^1\)). So horizontal asymptote is \(y = 0\) (the previous answer of \(y = 1\) was incorrect).

Step3: Find Vertical Asymptote

Vertical asymptotes occur where the denominator of the simplified function is zero (and the numerator is not zero there). For \(f(x)=\frac{4}{x}\), the denominator is zero when \(x = 0\). So vertical asymptote is \(x=0\).

Step4: Find x - Intercept

To find the x - intercept, set \(y = 0\) (i.e., \(f(x)=0\)) in the simplified function. But \(\frac{4}{x}=0\) has no solution because \(\frac{4}{x}\) can never be zero for any real \(x\). So there is no x - intercept.

Step5: Find y - Intercept

To find the y - intercept, set \(x = 0\) in the function. But in the simplified function \(f(x)=\frac{4}{x}\), \(x = 0\) is not in the domain (vertical asymptote at \(x = 0\)). Also, in the original function, when \(x = 0\), the denominator is zero, so there is no y - intercept.

Step6: Find Hole

A hole occurs at the value of \(x\) that makes both the numerator and denominator zero in the original function (before canceling). We had a common factor of \((x - 5)\), so when \(x=5\), both numerator and denominator are zero. To find the y - coordinate of the hole, substitute \(x = 5\) into the simplified function \(f(x)=\frac{4}{x}\), so \(f(5)=\frac{4}{5}\). So the hole is at \((5,\frac{4}{5})\).

Answer:

Horizontal Asymptote: \(y = 0\)
Vertical Asymptote: \(x = 0\)
x - Intercept: No x - intercept
y - Intercept: No y - intercept
Hole: \((5,\frac{4}{5})\)