Answer:
The domain is all real numbers
The y-intercept is 3.
Step-by-step explanation:
Let's analyze each statement separately. The function is
[tex]f(x)=(x+1)^2+2[/tex]
We have the following statements:
The domain is all real numbers, --> TRUE. The domain of a function is the set of values that the variable x can take. For this function, there are no restriction on the values that x can take, so the domain is all real numbers.
The range is all real numbers greater than or equal to 1 --> FALSE. The range of a function is the set of values that the variable y can take. For this function, we see that the factor [tex](x+1)^2[/tex] is always equal or greater than zero; this means that the minimum of the function is [tex]f(x)=2[/tex], so the range is all real numbers greater than or equal to 2.
The y-intercept is 3. --> TRUE. The y-intercept is the value of the function when x = 0. For this function, if we substitute x = 0, we find:
[tex]f(0)=(0+1)^2+2=1^2+2=3[/tex]
The graph of the function is 1 unit up and 2 units to the left from the graph of [tex]y=x^2[/tex] --> FALSE. The graph of a function is scaled n units up when [tex]g(x)=f(x)+n[/tex]; in this case, we see that the factor n in our fuction is n = 2, so the function is actually scaled 2 units up, not 1.
The graph has two x-intercepts --> FALSE. The x-intercept of a graph is the value of x when [tex]f(x)=0[/tex]. If we require [tex]f(x)=0[/tex] for our function, we get:
[tex]0=(x+1)^2+2\\\rightarrow (x+1)^2=-2[/tex]
However, this equation has no solutions: so, the graph has no x-intercepts.
