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Can you help me please

Can you help me please-example-1
User Sanderfish
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1 Answer

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Answer:

  • domain: x ∉ {-4, 3}
  • range: y ∉ {1}
  • horizontal asymptote: y=1
  • vertical asymptote: x=3

Explanation:

The expression reduces to ...


(x^2+9x+20)/(x^2+x-12)=((x+4)(x+5))/((x+4)(x-3))=(x+5)/(x-3)\quad (x\\e-4)

The domain is limited to values of x where the expression is defined. It is undefined where the denominator is zero, at x=-4 and x=3. The graph of the expression has a "hole" at x=4, where the numerator and denominator factors cancel.

  • the domain is all real numbers except -4 and +3

The function approaches the value of 1 as x gets large in magnitude, but it cannot take on the value of 1.

  • the range is all real numbers except 1

As discussed in 'range', there is a horizontal asymptote at y=1. That is the value you would get if you were to determine the quotient of the division:*

(x+5)/(x-3) = 1 + (8/(x-3)) . . . . quotient is 1

There is a vertical asymptote at the place where the denominator is zero in the simplified expression: x = 3.

  • vertical asymptote at x=3; horizontal asymptote at y=1

_____

* For some rational functions, the numerator has a higher degree than the denominator. In those cases, the quotient may be some function of x. The "end behavior" of the expression will match that function. (Sometimes it is a "slant asymptote", sometimes a higher-degree polynomial.)

Can you help me please-example-1
User Petrus Theron
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