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2 years ago

What is the axis of symmetry of the graph of f(x)=−2x²+8x−2?

Mathematics

1 answer:

FromTheMoon [43]2 years ago

7 0

For the given parabola, the axis of symmetry is x = 2.

<h3>How to get the axis of symmetry?</h3>

For any given parabola, we define the axis of symmetry as a line that divides the parabola in two equal halves.

For a regular parabola, we define the axis of symmetry as:

x = h

Where h is the x-component of the vertex.

Remember that for the general parabola:

y = a*x^2 + b*x + c

The x-value of the vertex is:

h = -b/(2a)

Then for the function:

f(x)=−2x²+8x−2

We get:

h = -8/(2*-2) = 2

Then the axis of symmetry is x = 2.

If you want to learn more about parabolas, you can read:

brainly.com/question/1480401

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Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to $(xyz)^2$ then it means that it is positive i.e $f(x,y,z)\geq 0$ .

Consider the function $F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)$

We want that the gradient of this function is equal to zero. That is (the calculations in between are omitted)

$\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0$

$\frac{\partial F}{\partial \lambda} = (x^2+y^2+z^2-1)=0$

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

$-2y\lambda = -2z\lambda$ . If lambda is not zero, then y =z. But, since $-2y\lambda=0$ and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set $y^2+z^2=1$ .

If both x,y are zero, then we have the solution set $z^2=1$ . We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

$\lambda = y^2z^2 = x^2z^2$ which implies $x^2=y^2$

Also, from the first and third equation we have that

$\lambda = y^2z^2 = x^2y^2$ which implies $x^2=z^2$

So, in this case, replacing this in the restriction we have $3z^2=1$ , which gives as another solution set. On this set, we have $x^2=y^2=z^2=\frac{1}{3}$ . Over this solution set, we have that the value of our function is $\frac{1}{3^3}= \frac{1}{27}$

4 0

3 years ago

Prove that every integer greater than 7 can be written by using 3's and 5's only. that is, for every n > 7 there exist non-ne

Taya2010 [7]

An integer may be a multiple of 3.
An integer may be 1 greater than a multiple of 3.
An integer may be 2 greater than a multiple of 3.

It is redundant to say an integer is 3 greater than a multiple of 3 (that's just a multiple of 3, we've got it covered). Same for 4, 5, 6, 7...

Let's consider a number which is a multiple of 3. Clearly, we can write 3+3+3+3+... until we reach the number. It can be written as only 3's.

Let's consider a number which is 2 greater than a multiple of 3. If we subtract 5 from that number, it'll be a multiple of 3. That means we can write the number as 5+3+3+3+3+... Of course, the number must be at least 8.

Let's consider a number which is 1 greater than a multiple of 3. If we subtract 5 from that number, it'll be 2 greater than a multiple of 3. If we subtract another 5, it'll be a multiple of 3. That means we can write the number as 5+5+3+3+3+3+... Of course, the number must be at least 13.

That's it. We considered all the numbers. We forgot 9, 10, 11, and 12, but these are easy peasy.

Beautiful question.

4 0

3 years ago

Given the graph below, which of the following statements is true?

nirvana33 [79]

Answer:

D) The graph does not represent a one-to-one function because the y-values between 0 to 2 are paired with multiple x-values.

Step-by-step explanation:

Let's find the ordered pairs.

(-4, 4), (-3, 2), (-2, 0), (-1, 0.5), (0, 1), (1, 1.5), (2, 2), (3, 0)

The graph passed through above points.

In the x-coordinates -3, 2 gives the same output. Therefore, the given function is not one-to-one.

Therefore, the answer D)

The graph does not represent a one-to-one function because the y-values between 0 to 2 are paired with multiple x-values.

Hope you will understand the concept.

Thank you.

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3 years ago

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1) C(2
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5) A(5+6=6+5
6) C(12x1=1x12
7) D(Exponents
8) A(addition and subtraction
9) A(55
10) B(19

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