Question

1. Identify the axis of symmetry for y = -3(x+3)^2-2. a. x = -2 b. x = 3 c. x = 2 d. x = -3 2. Choose the correct axis of symmetry for x = -4(y -4)^2+6 a. y = -6 b. y = -4 c. y =6 d. y = 4

Answer 1

Answer:

The answer is :

1. D
2. D

Step-by-step explanation:

Axis of symmetry is the equation where it cuts the middle of the quadratic graph.

For quadratic equation in the form of (x+a)² + b, the axis of symmetry will be (x+a) = 0 which is x = -a :

Question 1,

$(x + 3) = 0$

$x = - 3$

Question 2,

$(y - 4 )= 0$

$y = 4$

Answer 2

Answer:

$\boxed{x=-3} \\ \boxed{y=4}$

Step-by-step explanation:

Axis of symmetry is a line that cuts the parabola in half touching the vertex.

Quadratic forms ⇒ y = ax² + bx + c or x = ay² + by + c

Axis of symmetry ⇒ x = $\frac{-b}{2a}$ or y = $\frac{-b}{2a}$

First problem:

y = -3(x+3)²-2

Write in quadratic form ⇒ y = ax² + bx + c

y = -3(x² + 6x + 9) - 2

y = -3x² -18x - 27 - 2

y = -3x² -18x - 29

a = -3, b = -18

Find axis of symmetry.

$x= \frac{-b}{2a}$

$x=\frac{--18}{2(-3)}$

$x=\frac{18}{-6}=-3$

Second problem:

x = -4(y -4)² +6

Write in quadratic form ⇒ x = ay² + by + c

x = -4(y² - 18y + 16) + 6

x = -4y² + 32y - 64 + 6

x = -4y² + 32y - 58

a = -4, b = 32

Find axis of symmetry.

$y= \frac{-b}{2a}$

$y=\frac{-32}{2(-4)}$

$y=\frac{-32}{-8}=4$