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vagabundo [1.1K]
3 years ago
15

If $f(x)$ is a polynomial of degree 3, and $g(x)$ is a polynomial of degree 5, then what is the degree of polynomial $2f(x) + 4g

(x)$?
Mathematics
1 answer:
worty [1.4K]3 years ago
6 0

Answer:

Degree 5

Step-by-step explanation:

Given

Degree of f(x) = 3

Degree of g(x) = 5

Required

Degree of 2f(x) + 4g(x)

<h2>Analyzing both polynomials</h2><h3>f(x)</h3>

2f(x) means 2 * f(x)

Since 2 is a constant

Multiplying f(x) by 2 will result in a polynomial with a degree of 3

Hence 2f(x) has a degree of 3

<h3 /><h3>g(x)</h3>

4g(x) means 4 * g(x)

Since 4 is also a constant

Multiplying g(x) by 4 will result in a polynomial with a degree of 5

Hence 4g(x) has a degree of 5

Having said that;

When 2 polynomials of different degrees are added together, the degree of the result will be the higher degree of both polynomials;

This means that;

Adding a polynomial of degree 3 and another of degree 5 will result in a polynomial of degree 5

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Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = 3^{x + 2}

The asymptote is given as x → -∞, f(x) = 3^{x + 2} → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = 3^{0 + 2} = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = 5^{1  - x}

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = 5^{1  - 0} = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = 2^{x + 3} - 1

The asymptote is 2^{x + 3}  → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) =  2^{0 + 3} - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, 2^{x + 3} - 1 = 0

2^{x + 3} = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2

The asymptote is \left (\dfrac{1}{2} \right)^{x - 2} → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6

The y-intercept is (0, 6)

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In 3 years she will earn 1,800.

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