Question

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Consider the function f(x) = x^2-16/x+3. Determine all asymptotes of this function including horizontal, vertical, and oblique (slant).

please check:
What are the vertical asymptotes of the function f(x) = 5x+1/x^2+x-6? --> x can not = 2, x = -3 correct?

What are the zeros of the function f(x) = 3x^2-3x-6/3x-6? --> -2 correct?

Consider the function f(x) = x^2+x-12/x-3. Describe the graph of this function. Include all discontinuities, intercepts, and the basic shape of the graph. --> disc = -4? x int = (3,-4)? y int = 3? basic shape = straight line? correct?

Answer 1

f(x) = (x^2-16) / (x+3)
f(x) has no horizontal asymptote.
Vertical assymptote is x = -3
Dividing the numerator by the denominator gives x - 3 remainder -7. Therefore, the oblique assymptote is x - 3.


the vertical asymptotes of the function f(x) = (5x + 1) / (x^2 + x - 6) = (5x + 1)/(x - 2)(x + 3) is x = 2 and x = -3.

the zeros of the function f(x) = (3x^2 - 3x - 6 ) / (3x - 6) = (3x^2 + 3x - 6x - 6) / (3x - 6) = (3x(x + 1) - 6(x + 1)) / (3x - 6) = (3x - 6)(x + 1)/(3x - 6) = x + 1 is x = -1

the function f(x) = (x^2 + x - 12) / (x - 3) = (x - 3)(x + 4) / (x - 3) is --> disc = 3, x-int = (-4, 0), y-int = 4? basic shape = straight line