Question

1. find the vertex and the opening direction of the parabola, $y = -3x^2 - 12x - 7$.

a) $(-2, 5)$, down b) $(2, 19)$, down c) $(2, 5)$, up d) $(-2, 5)$, up
d
a
c
b

Explanation:

Step1: Find x-coordinate of vertex

For a quadratic \( y = ax^2 + bx + c \), vertex x-coordinate is \( -\frac{b}{2a} \). Here, \( a = -3 \), \( b = -12 \). So, \( -\frac{-12}{2(-3)} = -\frac{12}{6} = -2 \).

Step2: Find y-coordinate of vertex

Substitute \( x = -2 \) into \( y = -3x^2 -12x -7 \). \( y = -3(-2)^2 -12(-2) -7 = -12 + 24 -7 = 5 \). So vertex is \( (-2, 5) \).

Step3: Determine opening direction

Since \( a = -3 < 0 \), the parabola opens down.

Answer:

a) \((-2, 5)\), down