Question

PLEASE HELP!!!!

1) For the function f(x)= 3(x-1)^2, identify the vertex, domain and range.
A) The vertex is (1, 2), the domain is all real numbers, and the range is y >_ 2
B) The vertex is (1, 2), the domain is all real numbers and the range is y <_ 2
C) The vertex is ( -1, 2), the domain is all real numbers and the range is y >_ 2
D) The vertex is (-1, 2), the domain is all real numbers, and the range is y <_ 2


2) what is the equation of the following graph in vertex form? (PICTURE INCLUDED BELOW)
A) y= (x- 4)^2 - 4
B) y= (x + 4)^2 - 4
C) y= (x + 2)^2 + 6
D) y= (x + 2)^2 + 12

3) What is the equation of the following graph in vertex form?(PICTURE INCLUDED BELOW)
A) y= (x - 2)^2 + 1
B) y= (x -1)^2 + 2
C) y= (x +2)^2 +1
D) y= (x + 2)^2 -1


4) For the function f(x)= – (x + 1)^2 + 4, identify the vertex, domain, and range.
A)The vertex is ( –1, 4), the domain is all real numbers, and the domain is y >_ 4
B) The vertex is (–1, 4), the domain is all real numbers, and the range is y <_ 4
C) The vertex is (1, 4), the domain is all real numbers, and the range is y >_ 4
D) The vertex is (1, 4), the domain is all real numbers, and the range is y <_ 4

Answer 1

Option A is correct answer.

Explanation
`a(x-h)x^(2) +k`

If the given function is given as

f(x) =
`a(x-h)x^(2) +k`

then vertex is (h,k) and Its domain is set of all real numbers

and range is given to be y ≥k for a>0

So here in the question the equation is given as

`3(x-1)x^(2) +2`

on comparing the equation with given standard equation

we get a=3 which is greater than zero

h =1 and k=2

therefore vertex is (1,2) ,Domain is set of all real number

and range is y≥2

That is option A is correct!

Answer 2

QUESTION 1

The given function is

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

This function is of the form:

`f(x)=a(x-h)^2+k`, where
`V(h,k)` is the vertex of the function.

Hence the vertex is

`(1,2)`

The function is defined for all real values of
`x`. Hence the domain is all real numbers.

To find the range, we let

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

`\Rightarrow y-2=3(x-1)^2`

`\Rightarrow (y-2)/(3)=(x-1)^2`

`\Rightarrow sqrt{(y-2)/(3)}=x-1`

`\Rightarrow x=sqrt{(y-2)/(3)}+1`

`x` is defined for
`(y-2)/(3)\geq 0`

`x` is defined for y\geq 2[/tex]

The correct answer is A

QUESTION 2

Based on the description, I was able to picture the diagram as shown in the attachment.

This graph has the vertex
`(h,k)=(4,-4)`, Hence the equation is of the form:

`f(x)=a(x-h)^2+k`

The equation is

`y=(x-4)^2-4`

The correct answer is A

QUESTION 3

Based on the description, the graph has vertices (2,1)

Since this is a minimum graph;

The equation is of the form;

`f(x)=a(x-h)^2+k`, where
`a>\:0`.

Hence the equation is

`y=(x-2)^2+1`

The correct answer is A.

QUESTION 5

The given function is

`f(x)= -(x+1)^2+4`

This equation is of the form
`f(x)=a(x-h)^2+k` where
`V(-1,4)` is the vertex .

The function is defined for all real values of
`x`. Hence the domain is all real numbers.

To find the range, we let

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

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

`\Rightarrow 4-y=(x+1)^2`

`\Rightarrow √(4-y)=x+1`

`\Rightarrow x=√(4-y)-1`

`x` is defined for
`4-y\geq 0`

`\Rightarrow -y\geq -4`

`\Rightarrow y\le 4`

Hence the range is the range is
`y\le 4`

B) The vertex is (–1, 4), the domain is all real numbers, and the range is
`y\le 4`