Question

compare $g(x) = -3x^2 - 6x + 6$ to $f$, shown in the graph. which function has a greater maximum value?
choose the correct answer below.
a. $f(x)$; the $y$-coordinate of the vertex of $f(x)$ is greater than the $y$-coordinate of the vertex of $g(x)$.
b. $g(x)$; the $y$-coordinate of the vertex of $g(x)$ is greater than the $y$-coordinate of the vertex of $f(x)$.
c. $f(x)$; the $x$-coordinate of the vertex of $f(x)$ is greater than the $x$-coordinate of the vertex of $g(x)$.

Explanation:

Step1: Find vertex of \( g(x) \)

For a quadratic function \( ax^2 + bx + c \), vertex x - coordinate is \( -\frac{b}{2a} \). For \( g(x) = -3x^2 - 6x + 6 \), \( a = -3 \), \( b = -6 \).
\( x = -\frac{-6}{2\times(-3)} = -1 \)
Substitute \( x = -1 \) into \( g(x) \):
\( g(-1) = -3(-1)^2 - 6(-1) + 6 = -3 + 6 + 6 = 9 \)
So vertex of \( g(x) \) is \( (-1, 9) \), max value \( 9 \).

Step2: Analyze graph of \( f(x) \)

From the graph, the parabola \( f(x) \) has vertex (let's estimate from grid: x between -2 and 3, y - coordinate seems less than 9, e.g., maybe around 5 - 8? Wait, actually, looking at the grid, the peak of \( f(x) \) is below \( y = 9 \). So max of \( f(x) \) is less than 9.

Answer:

B. \( g(x) \); The y - coordinate of the vertex of \( g(x) \) is greater than the y - coordinate of the vertex of \( f(x) \).