Answer:
Option A is correct answer.
Step-by-step explanation[tex]a(x-h)x^{2} +k[/tex]
If the given function is given as
f(x) =[tex]a(x-h)x^{2} +k[/tex]
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
[tex]3(x-1)x^{2} +2[/tex]
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!
QUESTION 1
The given function is
[tex]f(x)=3(x-1)^2+2[/tex]
This function is of the form:
[tex]f(x)=a(x-h)^2+k[/tex], where [tex]V(h,k)[/tex] is the vertex of the function.
Hence the vertex is
[tex](1,2)[/tex]
The function is defined for all real values of [tex]x[/tex]. Hence the domain is all real numbers.
To find the range, we let
[tex]y=3(x-1)^2+2[/tex]
[tex]\Rightarrow y-2=3(x-1)^2[/tex]
[tex]\Rightarrow \frac{y-2}{3}=(x-1)^2[/tex]
[tex]\Rightarrow sqrt{\frac{y-2}{3}}=x-1[/tex]
[tex]\Rightarrow x=sqrt{\frac{y-2}{3}}+1[/tex]
[tex]x[/tex] is defined for [tex]\frac{y-2}{3}\geq 0[/tex]
[tex]x[/tex] 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 [tex](h,k)=(4,-4)[/tex], Hence the equation is of the form:
[tex]f(x)=a(x-h)^2+k[/tex]
The equation is
[tex]y=(x-4)^2-4[/tex]
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;
[tex]f(x)=a(x-h)^2+k[/tex], where [tex]a>\:0[/tex].
Hence the equation is
[tex]y=(x-2)^2+1[/tex]
The correct answer is A.
QUESTION 5
The given function is
[tex]f(x)= -(x+1)^2+4[/tex]
This equation is of the form [tex]f(x)=a(x-h)^2+k[/tex] where [tex]V(-1,4)[/tex] is the vertex .
The function is defined for all real values of [tex]x[/tex]. Hence the domain is all real numbers.
To find the range, we let
[tex]y=-(x+1)^2+4[/tex]
[tex]y-4=-(x+1)^2[/tex]
[tex]\Rightarrow 4-y=(x+1)^2[/tex]
[tex]\Rightarrow \sqrt{4-y}=x+1[/tex]
[tex]\Rightarrow x=\sqrt{4-y}-1[/tex]
[tex]x[/tex] is defined for [tex]4-y\geq 0[/tex]
[tex]\Rightarrow -y\geq -4[/tex]
[tex]\Rightarrow y\le 4[/tex]
Hence the range is the range is [tex]y\le 4[/tex]
B) The vertex is (–1, 4), the domain is all real numbers, and the range is [tex]y\le 4[/tex]
