Question

which of the following is the graph of this quadratic function?
y = -3x² - 6x + 2

Explanation:

Step1: Determine the direction of the parabola

For a quadratic function \( y = ax^2 + bx + c \), if \( a < 0 \), the parabola opens downward. Here, \( a=-3<0 \), so the parabola opens downward. This eliminates the third graph (which opens upward).

Step2: Find the vertex's x - coordinate

The x - coordinate of the vertex of a quadratic function \( y = ax^2+bx + c \) is given by \( x=-\frac{b}{2a} \). For \( y=-3x^{2}-6x + 2 \), \( a=-3 \) and \( b = - 6 \). Then \( x=-\frac{-6}{2\times(-3)}=-\frac{6}{6}=-1 \).

Step3: Analyze the y - intercept

The y - intercept of a quadratic function \( y = ax^2+bx + c \) is found by setting \( x = 0 \). When \( x = 0 \), \( y=-3(0)^{2}-6(0)+2=2 \). So the graph should pass through \( (0,2) \).

Now let's analyze the remaining two graphs (first and second):

• For the first graph: The vertex's x - coordinate seems to be around \( x = 0.5 \) (not - 1), so it does not match.
• For the second graph: The vertex's x - coordinate is \( x=-1 \) (since the axis of symmetry is \( x = - 1\)) and it passes through \( (0,2) \) (the y - intercept is 2) and it opens downward.

Answer:

The second graph (the one with x - intercepts around \( x=-2 \) and \( x = 0 \), vertex at \( x=-1 \), y - intercept at \( y = 2 \) and opening downward)