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Natasha2012 [34]
2 years ago
10

The function f(x) = x3 – 8x2 + x + 42 has zeros located at 7, –2, 3. Verify the zeros of f(x) and explain how you verified them.

Describe the end behavior of the function.
Mathematics
1 answer:
xeze [42]2 years ago
3 0

Answer:

  • zeros are {-2, 3, 7} as verified by graphing
  • end behavior: f(x) tends toward infinity with the same sign as x

Step-by-step explanation:

A graphing calculator makes finding or verifying the zeros of a polynomial function as simple as typing the function into the input box.

<h3>Zeros</h3>

The attachment shows the function zeros to be x ∈ {-2, 3, 7}, as required.

<h3>End behavior</h3>

The leading coefficient of this odd-degree polynomial is positive, so the value of f(x) tends toward infinity of the same sign as x when the magnitude of x tends toward infinity.

  • x → -∞; f(x) → -∞
  • x → ∞; f(x) → ∞

__

<em>Additional comment</em>

The function is entered in the graphing calculator input box in "Horner form," which is also a convenient form for hand-evaluation of the function.

We know the x^2 coefficient is the opposite of the sum of the zeros:

  -(7 +(-2) +3) = -8 . . . . x^2 coefficient

And we know the constant is the opposite of the product of the zeros:

  -(7)(-2)(3) = 42 . . . . . constant

These checks lend further confidence that the zeros are those given.

(The constant is the opposite of the product of zeros only for odd-degree polynomials. For even-degree polynomials. the constant is the product of zeros.)

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A sixth grader weighs 90 pounds, which is 120% of what he weighed in fourth grade. How much did he weigh in fourth grade?
AlekseyPX
90=1.2f
(f stands for how much he weighed in fourth grade)
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So, divide by 1.2 on both sides.
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3 years ago
In Hamilton County, Ohio the mean number of days needed to sell a home is days (Cincinnati Multiple Listing Service, April, 2012
Diano4ka-milaya [45]

Answer:

t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974    

p_v =2*P(t_{39}

If we compare the p value with a significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.    

Step-by-step explanation:

Assuming the following problem: "In Hamilton County, Ohio the mean number of days needed to sell a home is 86 days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale 40 of homes in a nearby county showed a sample mean of 80 days with a sample standard deviation of 20 days. Conduct a hypotheses test to determine whether the mean number of days until a home is sold is different than the Hamilton county mean of 86 days in the nearby county. "

1) Data given and notation    

\bar X=80 represent the sample mean

s=20 represent the sample standard deviation

n=40 sample size    

\mu_o =86 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean is different from 86, the system of hypothesis would be:    

Null hypothesis:\mu = 86    

Alternative hypothesis:\mu \neq 86    

We don't know the population deviation, so for this case we can use the t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic    

We can replace in formula (1) the info given like this:    

t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974    

4) Calculate the P-value    

First we need to find the degrees of freedom given by:

df=n-1=40-1=39

Since is a two tailed test the p value would be:    

p_v =2*P(t_{39}

In Excel we can use the following formula to find the p value "=2*T.DIST(-1.8974,39,TRUE)"  

5) Conclusion    

If we compare the p value with a significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.    

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ANSWER:

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On solving eq(1), we get

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Plug the value of x in eq(2)

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So, given system of equations is not correct.

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