Question

Hello plz help with these two questions ty

Answer 1

I'll do problem 1 to get you started off.

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Problem 1, part A

The term "root" is another way of saying "zeros of a function".

The given roots are: 8, -1, -3

If x = k is a root, then x-k is a factor. Based on this rule, we have these factors:

• x-8
• x+1
• x+3

So we would have this expansion

(x-8)(x+1)(x+3)

(x-8)(x^2+4x+3)

x(x^2+4x+3) - 8(x^2+4x+3)

x^3+4x^2+3x - 8x^2-32x-24

x^3 - 4x^2 - 29x - 24

Answer: f(x) = x^3 - 4x^2 - 29x - 24

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Problem 1, part B

Factor x^2-x-2 to get (x-2)(x+1)

Note that we have (x+1) show up, which was also a factor in f(x). Refer to part A.

What this means is that when dividing f(x) over x^2-x-2, we'll have the pair of (x+1) factors cancel out.

`g(x) = (f(x))/(x^2-x-2)\\\\g(x) = ((x-8)(x+1)(x+3))/((x-2)(x+1))\\\\g(x) = ((x-8)(x+3))/(x-2)\\\\g(x) = (x^2-5x-24)/(x-2)\\\\`

To find the slant asymptote of g(x), we'll need to do synthetic division as shown in the diagram below. Specifically refer to the diagram labeled "problem 1, part B". Polynomial long division is another option, but it's more cumbersome in my opinion.

After doing the synthetic division, we see that x-3 is the quotient and -30 is the remainder. We ignore the remainder and focus on the quotient.

The quotient tells us that the slant asymptote is the straight line y = x-3

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Problem 1, part C

The discontinuities of g(x) will occur whenever the denominator x^2-x-2 is equal to zero.

x^2-x-2 = 0

(x-2)(x+1) = 0

x-2 = 0 or x+1 = 0

x = 2 or x = -1

Because the (x+1) factors canceled, this means that x = -1 is a removable discontinuity. It's a single point removed or an open hole at this location on the curve. Refer to the diagram labeled "problem 1, part C" below.

In that same diagram, we have a vertical asymptote at x = 2 because the factor (x-2) in the denominator does not cancel out. This is an infinite discontinuity due to the fact that the curve is approaching plus or minus infinity whether we approach the vertical asymptote from the left or right sides respectively.

Take note that the slant asymptote, aka oblique asymptote, is graphed as well. It's what the curve approaches as x goes off to infinity. The presence of an oblique asymptote means we don't have a horizontal asymptote.

Answer 2

9514 1404 393

Answer:

1A: f(x) = x^3 -4x^2 -29x -24

1B: g(x) = (x -8)(x +3)/(x -2); slant asymptote: y = x-3

1C: vertical asymptote: x = 2; hole: (-1, 6)

2A: 0 ≤ t ≤5

2B: 5 mg/L

Explanation:

1.

A: The factored form of the function is ...

f(x) = (x -8)(x +1)(x +3)

Multiplying this out, we get ...

f(x) = (x -8)(x² +4x +3) = x(x² +4x +3) -8(x² +4x +3) = x³ +4x² +3x -8x² -32x -24

f(x) = x³ -4x² -29x -24

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B:

`g(x)=((x-8)(x+1)(x+3))/(x^2-x-2)=((x-8)(x+1)(x+3))/((x+1)(x-2))\\\\\boxed{g(x)=((x-8)(x+3))/(x-2)}\qquad x\\e-1\\\\g(x)=(x -3)-(30)/(x-2)\\\\\text{slant aymptote: }y=x-3`

The first attachment shows the synthetic division that tells us the slant asymptote.

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C: There is a vertical asymptote at x = 2 (where the denominator of g(x) is zero. There is a "hole" at (x, y) = (-1, 6), where the function g(x) is undefined (the denominator is zero).

The graph in the second attachment shows the asymptotes and hole in g(x).

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2.

A: The denominator is always positive for t ≥ 0, so the reasonable domain will be governed by the numerator of C(t). That factors as (-10t)(t -5), which is only non-negative for 0 ≤ t ≤ 5. The reasonable domain of C(t) is 0 ≤ t ≤ 5.

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B. The graph in the 3rd attachment shows the greatest concentration will be 5 mg/L after 1 hour.