For the function f(x) = 3(x − 1)2 + 2, identify the vertex, domain, and range.

Ok, our function is:

f(x) = 3*(x - 1)^2 + 2.

First, domain:

We should assume that the domain is all the set of real numbers, and then we see if for some value we have a problem.

In this case we do not see any problem (we can not have a zero in the denominator, and there is no function that has problems with some values of x)

Then the domain is the set of all real numers.

Vertex:

Let's expand our function:

f(x) = 3*x^2 - 3*2*x + 1 + 2

f(x) = 3*x^2 -6*x + 2

The vertex of a quadratic function:

a*x^2 + b*x + c is at:

x = -b/2a

here we have:

a = 3 and b = -6

x = 6/2*3 = 6/6 = 1.

And the value of y at that point is:

f(1) = 3*(1 - 1)^2 + 2 = 2

Then the vertex is at: (1, 2)

Range:

The range is the set of all the possible values of y.

Ok, we can see that the leading coefficient is positive, this means that the arms of our quadratic function will go up.

Then the minimal value of our quadratic function is the value at the vertex, y = 2.

This means that the range can be written as:

R = y ≥ 2

So the range is the set of all real numbers that are larger or equal than 2.

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.