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plot all of the existing five features of the following rational functi…

Question

plot all of the existing five features of the following rational function (some may not be needed). if you get a fraction or decimal then plot as close to the true location as possible. $f(x) = \frac{-2x - 4}{x^2 - 4}$ plot rational function vertical asymptote horizontal asymptote x-intercept y-intercept hole click on a feature then drag it into place.

Explanation:

Step1: Simplify the function

First, factor the numerator and the denominator:

  • Numerator: \(-2x - 4 = -2(x + 2)\)
  • Denominator: \(x^2 - 4 = (x - 2)(x + 2)\)

So, \(f(x)=\frac{-2(x + 2)}{(x - 2)(x + 2)}\). We can cancel out the common factor \((x + 2)\) (note that \(x
eq - 2\) to avoid division by zero), so the simplified function is \(f(x)=\frac{-2}{x - 2}\) with a hole at \(x=-2\) (since \((x + 2)\) was a common factor).

Step2: Find Vertical Asymptote

Vertical asymptotes occur where the denominator of the simplified function is zero (and the numerator is not zero there). For \(f(x)=\frac{-2}{x - 2}\), set \(x - 2=0\), so \(x = 2\) is the vertical asymptote.

Step3: Find Horizontal Asymptote

For rational functions, if the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\). Here, the degree of the numerator (after canceling) is \(0\) and the degree of the denominator is \(1\), so horizontal asymptote is \(y = 0\).

Step4: Find x - Intercept

To find the x - intercept, set \(y = f(x)=0\). For \(f(x)=\frac{-2}{x - 2}\), \(\frac{-2}{x - 2}=0\) has no solution because the numerator is \(-2
eq0\). So, there is no x - intercept.

Step5: Find y - Intercept

To find the y - intercept, set \(x = 0\) in the original function (before canceling, to check for the hole and intercept).
\(f(0)=\frac{-2(0)-4}{0^2 - 4}=\frac{-4}{-4}=1\). But we also need to check the hole. The hole is at \(x=-2\), when \(x = 0\), we can use the simplified function \(f(0)=\frac{-2}{0 - 2}=\frac{-2}{-2}=1\). So the y - intercept is at \((0,1)\).

Step6: Find Hole

The hole occurs at the value of \(x\) where the common factor was canceled, i.e., \(x=-2\). To find the y - coordinate of the hole, substitute \(x = - 2\) into the simplified function (or the original function before canceling, but simplified is easier). Using \(f(x)=\frac{-2}{x - 2}\), when \(x=-2\), \(f(-2)=\frac{-2}{-2 - 2}=\frac{-2}{-4}=\frac{1}{2}\). So the hole is at \((-2,\frac{1}{2})\).

Answer:

  • Vertical Asymptote: \(x = 2\)
  • Horizontal Asymptote: \(y = 0\)
  • y - Intercept: \((0,1)\)
  • Hole: \((-2,\frac{1}{2})\)
  • No x - Intercept