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nexus9112 [7]
2 years ago
14

Can you help me create a polynomial function matching the criteria on the left please and thanks

Mathematics
2 answers:
Olenka [21]2 years ago
7 0

Answer:

f(x) = \dfrac{1}{8}(x + 4)^2(x + 1)(x - 3)

Step-by-step explanation:

From inspection of the graph, we can see that the curve intercepts the x-axis at (-4, 0), (-1, 0) and (3, 0)

Therefore,

x = -4 ⇒ x + 4 = 0

x = -1 ⇒ x + 1 = 0

x = 3 ⇒ x - 3 = 0

Because (-4, 0) touches the x-axis, then (x + 4)² will be a factor

So (x + 4)², (x + 1) and (x - 3) are all factors of the polynomial

\implies f(x) = \dfrac{1}{n}(x + 4)^2(x + 1)(x - 3)

If we multiply the constants, this will give us the y-intercept:

⇒ 4² x 1 x -3 = -48

From inspection of the graph, the y-intercept is -6

So to get from -48 to -6 we need to multiply -48 by 1/8

Therefore, n = 1/8

\implies f(x) = \dfrac{1}{8}(x + 4)^2(x + 1)(x - 3)

dybincka [34]2 years ago
3 0

Step-by-step explanation:

the factored representation has at least one major advantage : the zero-solutions are directly visible.

as whenever one factor turns zero, the whole function turns zero.

the zero solutions are

(-4, 0)

(-1, 0)

(3, 0)

based on this we get

(x + 4)(x + 1)(x - 3)

for x = -4 the first factor is 0.

for x = -1 the second factor is 0.

for x = 3 the third factor is zero.

the y-intercept gives us a hint about 1/#, as this is the y value when x = 0.

this is then

1/# × (0+4)(0+1)(0-3) = 1/# × -12

the point (0, -6) tells us this must be equal to -6.

1/# × -12 = -6

1/# = -6/-12 = 1/2

# = 2

so, we get

f(x) = 1/2 × (x + 4)(x + 1)(x - 3)

but that is still not the whole result.

why ?

because the first zero solution (-4, 0) is actually the merger of 2 zero solutions, as the vertex of a loop is only touching the x-axis.

if we shift the whole graph a little bit down, we see that this x-intercept would turn into 2 intercepts.

so, this function has actually 4 zero solutions.

that tells us that the polynomial is of 4th degree (there must be a "x⁴" somewhere in the expression as highest exponent of x).

but multiplying only 3 factors based on x will give us only x³ as highest x exponent.

the solution : the first factor has to represent 2 zero solutions as the graph indicates.

how ?

by squaring that factor.

so, our function looks like

f(x) = 1/2 × (x + 4)²(x + 1)(x - 3)

now we get "x⁴" in the expression.

but wait, now for x = 0 the 1/# is not right anymore.

we have now

1/# × (0+4)²(0+1)(0-3) = 1/# × 16×1×-3 = 1/# × -48

so,

1/# × -48 = -6

1/# = -6/-48 = 1/8

# = 8

so, finally, the fully correct function is

f(x) = 1/8 × (x + 4)²(x + 1)(x - 3)

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Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

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The height in centimeters of seedlings treated with Herbicide B, \overline x _2 = 109.1 cm

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The 90% confidence interval for μ₂ - μ₁, is given as follows;

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The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

t_{(0.025, \, 18)} = 1.734

C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

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Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

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The two sample t-statistic is given as follows;

t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}

t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147

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b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

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When x < 0, one possible value is x = -1, hence evaluating the functions at these values:

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Quadrant I      -          0° to 90°

Quadrant II     -       90° to 180°

Quadrant III    -      180° to 270°

Quadrant IV    -      270° to 360°

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Quadrant III   -       -90° to -180°

Quadrant II    -      -180° to -270°

Quadrant I     -      -270° to -360°

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