Which functions have removable discontinuities (holes)? Check all of the boxes that apply.

Mathematics

2 answers:

Naya [18.7K]3 years ago

4 0

Answer:

a b d

Step-by-step explanation:

Zolol [24]3 years ago

3 0

Answer:

Option A,B,D

Step-by-step explanation:

For f(x) = (x-1)/(x²-1)

f(x) = (x-1)/(x-1)(x+1)

Therefore hole at x = 1

For f(x) = (x²-9)/(x²+7x+12) = (x-3)(x+3)/(x²+4x+3x+12) = (x-3)(x+3)/(x+3)(x+4)

Common factor (x+3) in numerator and denominator so

Function has hole at x = -3

For f(x) = (x² + 4x + 4)/(x² + 2x - 8) = (x+2)²/(x² +4x -2x -8) = (x+2)²/(x+4)(x-2)

This function has no discontinuity.

For f(x) = (x +7)/(x²+5x-14) = (x+7)/(x² + 7x -2x -14) = (x+7)/(x-2)(x+7)

This function has a hole at x = -7

Option A,B,D have the removable discontinuity.

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Delicious77 [7]

Answer:

$f(x)=-a(x+2)(2x-1)(x-3)$ where a>0.

To graph the the polynomial, begin in the left top of quadrant 2. Then draw downwards to the first real zero on the x-axis at -2. Cross the x-axis and then curve back up to 1/2 on the x-axis. Cross through again and curve back down to cross for the last time at 3 on the x-axis. The graph then ends going down towards the right in quadrant 4. It forms an s shape.

Step-by-step explanation:

The real zeros are the result of setting each factor of the polynomial to zero. By reversing this process, we find:

zero -2 is factor (x+2)

zero 1/2 is factor (2x-1)

zero 3 is factor (x-3)

We write them together with an unknown leading coefficient a which is negative so -a.

$f(x)=-a(x+2)(2x-1)(x-3)$ where a>0

The leading coefficient of a polynomial determines the direction of the graph's end behavior.

A positive leading coefficient has the end behavior point up when an even degree and point opposite directions when an odd degree with the left down and the right up.

A negative leading coefficient has the end behavior point down when an even degree and point opposite directions when an odd degree with the left up and the right down.