1. f(x) = (x + 2)2 + 1
direction of opening:
vertex:
axis of symmetry:
y-intercept:
buddy point:
what is the direction of opening

1) Direction of opening: upward

2) Vertex is the point (-2,1)

3) Axis of Symmetry: x=-2

4) y-intercept: point (0,5)

5) Buddy point:  and  

Step-by-step explanation:

we have

This is the equation of a vertical parabola in vertex form

The general equation of a vertical parabola in vertex form is equal to

where

(h,k) is the vertex

a is the coefficient of variable x^2

If a > 0 -----> the parabola open upward and the vertex is a minimum

If a < 0 -----> the parabola open downward and the vertex is a maximum

The axis of symmetry is equal to the x-coordinate of the vertex

x=h

In this problem we have

The vertex is the point (-2,1)

The coefficient a is equal to a=1

so

a> 0

the parabola open upward and the vertex is a minimum

The axis of symmetry is

x=-2

Find out the y-intercept

Remember that the y-intercept is the value of y when the value of x is equal to zero

For x=0

The y-intercept is the point (0,5)

Find the x-intercepts

Remember that the x-intercept is the value of x when the value of y is equal to zero

so

For f(x)=0

The roots are complex

take square root both sides

Remember that

The solutions are

therefore

1) Direction of opening: upward

2) Vertex is the point (-2,1)

3) Axis of Symmetry: x=-2

4) y-intercept: point (0,5)

5) Buddy point:  and  

(x , y) → (-x , y), (x , y) → (x + 1 , y + 1) ⇒ answer B

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis  then its image is (x , -y)

- If point (x , y) reflected across the y-axis  then its image is (-x , y)

- If the point (x , y) translated horizontally to the right by h units

 then its image is (x + h , y)

- If the point (x , y) translated horizontally to the left by h units

 then its image is (x - h , y)

- If the point (x , y) translated vertically up by k units

 then its image is (x , y + k)

- If the point (x , y) translated vertically down by k units

 then its image is (x , y - k)

* Lets solve the problem

∵ The vertices of the Δ BCD are:

# B = (-3 , 0)

# C = (2 , -1)

# D = (-1 , 2)

∵ The images of its vertices after a series of transformations are:

# B″ = (4 , 1)

# C" = (-1 , 0)

# D" = (2 , 3)

* Lets find the write answer

- (x , -y) means Δ BCD across x-axis

∴ B' = (-3 , 0)

∴ C' = (2 , 1)

∴ D' = (-1 , -2)

- (x + 1 , y + 1) means translate ΔB'C'D' 1 unit to the right and 1 unit up

∴ B" = (-2 , 1)

∴ C" = (0 , 1)

∴ D" = (3 , 4)

- Not the vertices of Δ B"C"D"

∴ Answer A is not our answer

- (-x , y) means reflect ΔBCD across y-axis

∴ B' = (3 , 0)

∴ C' = (-2 , -1)

∴ D' = (1 , 2)

- (x + 1 , y + 1) means translate ΔB'C'D' 1 unit to the right and 1 unit up

∴ B" = (4 , 1)

∴ C" = (-1 , 0)

∴ D" = (2 , 3)

- The same with the vertices of Δ B"C"D"

∴ Answer B is our answer

∴ series of transformations would produce ΔB"C"D" is

   (x , y) → (-x , y), (x , y) → (x + 1 , y + 1)