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3 years ago

Find the vertex of the parabola formed by the following quadratic function: y = 3x^2 – 6x + 2

1 answer:

7 0

Vertex form is y=a(x-h)^2+k, so we can rearrange to that form...

y=3x^2-6x+2 subtract 2 from both sides

y-2=3x^2-6x divide both sides by 3

(y-2)/3=x^2-2x, halve the linear coefficient, square it, add it to both sides...in this case: (-2/2)^2=1 so

(y-2)/3+1=x^2-2x+1 now the right side is a perfect square

(y-2+3)/3=(x-1)^2

(y+1)/3=(x-1)^2 multiply both sides by 3

y+1=3(x-1)^2 subtract 1 from both sides

y=3(x-1)^2-1 so the vertex is:

(1, -1)

...

Now if you'd like you can commit to memory the vertex point for any parabola so you don't have to do the calculations like what we did above. The vertex of any quadratic (parabola), ax^2+bx+c is:

x= -b/(2a), y= (4ac-b^2)/(4a)

Then you will always be able to do a quick calculation of the vertex :)

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A college infirmary conducted an experiment to determine the degree of relief provided by three cough remedies. Each cough remed

KiRa [710]

Answer:

$p_v = P(\chi^2_{4,0.05} >3.81)=0.43233$

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed. So we can say that the 3 remedies ar approximately equally effective.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

NyQuil Robitussin Triaminic Total

No relief 11 13 9 33

Some relief 32 28 27 87

Total relief 7 9 14 30

Total 50 50 50 150

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the three remedies

H1: There is a difference in the three remedies

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{50*33}{150}=11$

$E_{2} =\frac{50*33}{150}=11$

$E_{3} =\frac{50*33}{150}=11$

$E_{4} =\frac{50*87}{150}=29$

$E_{5} =\frac{50*87}{150}=29$

$E_{6} =\frac{50*87}{150}=29$

$E_{7} =\frac{50*30}{150}=10$

$E_{8} =\frac{50*30}{150}=10$

$E_{9} =\frac{50*30}{150}=10$

And the expected values are given by:

NyQuil Robitussin Triaminic Total

No relief 11 11 11 33

Some relief 29 29 29 87

Total relief 10 10 10 30

Total 50 50 50 150

And now we can calculate the statistic:

$\chi^2 = \frac{(11-11)^2}{11}+\frac{(13-11)^2}{11}+\frac{(9-11)^2}{11}+\frac{(32-29)^2}{29}+\frac{(28-29)^2}{29}+\frac{(27-29)^2}{29}+\frac{(7-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(14-10)^2}{10} =3.81$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(3-1)=4$

And we can calculate the p value given by:

$p_v = P(\chi^2_{4,0.05} >3.81)=0.43233$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.81,4,TRUE)"

4 0

3 years ago

To find 10, Beau found 32 = 9 and 42 = 16. He said that since 10 is between 9 and 16, 10 is between 3

AlexFokin [52]

Given that:

Consider it is $\sqrt{10}$ instead of 10 on two places.

$\sqrt{10}$ is between 3 and 4. So, Beau thinks a good estimate for $\sqrt{10}$ is = 3.5.

Solution:

To find $\sqrt{10}$ , Beau found 3² = 9 and 4² = 16.

He said that since 10 is between 9 and 16.

Since 10 is close to 9, therefore $\sqrt{10}$ must be close to 3. So, Beau's estimate is high.

Now,

$(3.1)^2=9.61$

$(3.2)^2=10.24$

Since, 10 lies between 9.61 and 10.24, therefore $\sqrt{10}$ must be lies between 3.1 and 3.2.

$\dfrac{3.1+3.2}{2}=3.15$

Therefore, the estimated value of $\sqrt{10}$ is 3.15.

4 0

3 years ago

The current cost of a typical desktop computer is $745. The cost in year 1983 was $1,445 with fewer capabilities. What is the CP

Fantom [35]

$ 700. seria
la respuesta

7 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

kykrilka [37]

The answer is B
$\frac{5 \sqrt{13} }{13}$

3 0

2 years ago

PLEASE HELP I NEED THIS ASAP ! 100 POINTS!!

Finger [1]

Answer:

Part A: The population are the American adult citizens, excluding the ones from Alaska and Hawaii. The population is the people which the sample is trying to represent, as a hole.

The sample is a portion of this population, and in this case is represented by a randomly selected amount of people whose response to the interview has been selected.

Part B: The sample here has been selected in two steps. The first step is the one that we must pay attention to: the numbered list of one million responses from around the US (excluding Alaska and Hawaii). Because these responses were obtained by an online survey, the sample looks like a convenience sampling, as it depends on the availability and willingness from participants to take part of the study (is not compulsory for everyone, so not everyone is going to response, then there are people that is not going to be represented by). The second step is the random selection of a part of the previous responses. This last part will ensure that, the individuals that took part of the group that was interviewed, are well represented in the results.

Part C: As it was mentioned, there is a selection bias, because the information from the sample comes from a specific group of people that has certain features that may not represent all American adults citizens. For example, the opinion of those people who do not use internet, will not be considered (and they may be a large number of persons). This situations weaken the conclusions obtained in the study, as they are not representative of the hole population.

Step-by-step explanation:

6 0

1 year ago