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Using the following function, explain how the following features (asymptotes,
intercepts, holes) can be identified/constructed in the graph.
3(2-3)(x+2)(x-4)/8(6+9) (2-3)

Using the following function explain how the following features asymptotes intercepts holes can be identifiedconstructed in the graph 323x2x4869 23 class=

Respuesta :

The given asymptote, intercepts and holes of the given function depends

on the numerator and denominator of the function from which we have;

  • The vertical asymptote is the line x = -9
  • The x-intercept occurs at the points, x = -2, and x = 4, the y-intercept is located at the point at which y = [tex]-\dfrac{1}{3}[/tex]

  • The hole is at the point (3, 0)

How can the properties of the function be found?

The given function is presented as follows;

[tex]f(x) = \mathbf{ \dfrac{3 \cdot (x - 3) \cdot (x + 2) \cdot (x - 4)}{8 \cdot (x + 9) \cdot (x - 3)}}[/tex]

Given that there is a common factor of (x - 3) in the numerator and the denominator,  we have;

[tex]f(x) = \mathbf{\dfrac{3 \cdot (x + 2) \cdot (x - 4)}{8 \cdot (x + 9)}}[/tex]

The asymptote is obtained by equating the denominator to zero as follows;

8·(x + 9) = 0

x = -9

  • The asymptote is the line x = -9

The intercepts is given by the points the x or y-value are zero, which gives;

[tex]At \ an \ intercept, \ f(x) = \dfrac{3 \cdot (x - 3) \cdot (x + 2) \cdot (x - 4)}{8 \cdot (x + 9) \cdot (x - 3)}= 0[/tex]

3·(x - 3)·(x + 2)·(x - 4) = 0

Therefore;

  • x = -2, x = 4 at the x-intercepts

[tex]At \ the \ \mathbf{y-intercept}, \ f(0) = \dfrac{3 \cdot (0- 3) \cdot (0 + 2) \cdot (0 - 4)}{8 \cdot (0 + 9) \cdot (0 - 3)}= -\dfrac{1}{3}[/tex]

The hole is given by the point where the function is zero.

Placing (x - 3) = 0 in the denominator

Therefore;

  • A hole is the point x = 3

Learn more about asymptotes and intercepts here:

https://brainly.com/question/9170705

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