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3 years ago

What is the equation in vertex form of the quadratic function with a vertex at (-1, -4) that goes through (1, 8)?

Mathematics

1 answer:

cestrela7 [59]3 years ago

5 0

Answer:

y = 3(x+1)^2 - 4

Step-by-step explanation:The general form of the equation of a quadratic function whose vertex is (h,k) and whose leading coefficient is a is:

y - k = a(x-h)^2, or

y = a(x-h)^2 - k

Substituting the coefficients of the vertex (-1, -4), we get:

y = a(x + 1)^2 - 4

Substituting the coordinates of the given point, (1,8), we get:

8 = a(1+1)^2 - 4, which simplifies to:

8 = a(2)^2 - 4, or

8 = 4a - 4. Then 4a = 12, and a = 3.

Thus, the desired equation is y = 3(x+1)^2 - 4 (answer j).

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F(x)=3/x+2-√x-3 f(19)=

77julia77 [94]

Answer:

- 27/7

Step-by-step explanation:

Given

f(x)=3/(x+2) - √(x-3)

To find

f(19)

Solution

f(19) = 3/(19 + 2) - √(19 - 3) = 3/21 - √16 = 1/7 - 4 = - 27/7

3 0

2 years ago

Change -2-3 into addition problem<br><img src="https://tex.z-dn.net/?f=%28%20-%202%20-%203%20%3D%20" id="TexFormula1" title="( -

SashulF [63]

Answer: - (2 + 3) = -5

<u>Step-by-step explanation:</u>

When combining numbers that have the same sign, you add them and then include the sign.

For example:

-2 - 3 = -(2 + 3) = -5

-6 - 5 = -(6 + 5) = -11

-7 - 8 = -(7 + 8) = -15

8 0

3 years ago

Read 2 more answers

Assume that the function f is a one-to-one function, (a) f(8) = 9, find f ^ -1 (9)

Annette [7]

(a) since we are given that f(8) = 9, the inverse of 9 is simply 8. Therefore, f^-1(9) = 8.

(b) Again, f is the inverse of f^-1, therefore, you simply switch numbers, and you get that f(-3) = -7.

8 0

3 years ago

What is the value of x: -7x = -2x + 15

Art [367]

Answer:

-3

Step-by-step explanation:

-7x=-2x+15

collect like terms

-7x+2x=15

-5x=15

divide both sides by -5

x=15/-5

x=-3

6 0

3 years ago

Read 2 more answers

For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df

rjkz [21]

Answer:

a) For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

b) For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

Part b

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

3 0

3 years ago