Question

PLEASE HELP!!!!

Answer 1

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

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

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

$\Rightarrow \frac{y-2}{3}=(x-1)^2$

$\Rightarrow sqrt{\frac{y-2}{3}}=x-1$

$\Rightarrow x=sqrt{\frac{y-2}{3}}+1$

$x$ is defined for $\frac{y-2}{3}\geq 0$

$x$ 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

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

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

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

$\Rightarrow \sqrt{4-y}=x+1$

$\Rightarrow x=\sqrt{4-y}-1$

$x$ 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 $y\le 4$

Answer 2

Answer:

Option A is correct answer.

Step-by-step 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!