Question

select the correct answer.
the table represents quadratic function g. which statement is true about the function?

x	-5	-4	-3	-2	-1	0
g(x)	-1	0	-1	-4	-9	-16

a. the maximum occurs at the function’s x-intercept.
b. the minimum occurs at the function’s y-intercept.
c. the minimum occurs at the function’s x-intercept.
d. the maximum occurs at the function’s y-intercept.

Explanation:

Step1: Analyze the symmetry of the quadratic function

For a quadratic function, the axis of symmetry is the vertical line that passes through the vertex. Looking at the table, when \( x = -5 \), \( g(x)=-1 \); when \( x = -3 \), \( g(x)=-1 \). The midpoint of \( x=-5 \) and \( x = -3 \) is \( x=\frac{-5 + (-3)}{2}=-4 \). So the axis of symmetry is \( x=-4 \).

Step2: Determine if it's a maximum or minimum

The values of \( g(x) \) around \( x = -4 \): for \( x=-5 \), \( g(x)=-1 \); \( x=-4 \), \( g(x)=0 \); \( x=-3 \), \( g(x)=-1 \); as \( x \) moves away from \( x=-4 \) (in both directions), \( g(x) \) decreases (e.g., \( x=-2 \), \( g(x)=-4 \); \( x=-1 \), \( g(x)=-9 \); \( x = 0 \), \( g(x)=-16 \)). So the function has a maximum at \( x=-4 \) (since the parabola opens downward, as the values decrease on both sides of \( x=-4 \)).

Step3: Find x - intercept and y - intercept

• x - intercept: The x - intercept is where \( g(x)=0 \). From the table, when \( x=-4 \), \( g(x)=0 \). So the x - intercept is at \( x=-4 \), \( y = 0 \).
• y - intercept: The y - intercept is where \( x = 0 \). From the table, when \( x = 0 \), \( g(x)=-16 \). So the y - intercept is at \( x = 0 \), \( y=-16 \).

Step4: Evaluate each option

• Option A: The maximum occurs at \( x=-4 \), which is the x - intercept (since at \( x=-4 \), \( g(x)=0 \), so it's the x - intercept). This seems correct. Let's check others.
• Option B: The minimum? But we saw the function has a maximum at \( x=-4 \), and as \( x \) moves away, \( g(x) \) decreases, so there is no minimum (or the minimum would be at \( \pm\infty \) if we consider the end - behavior, but from the table, the values are decreasing). Also, the y - intercept is at \( (0,-16) \), which is not a minimum (since values keep decreasing as \( x \) increases beyond 0? Wait, no, for a quadratic function \( ax^{2}+bx + c \), if the coefficient of \( x^{2} \) is negative, it opens downward, so it has a maximum. Let's confirm the quadratic function. Let's assume \( g(x)=a(x + 4)^{2}+k \). Using \( x=-5 \), \( g(-5)=-1 \): \( a(-5 + 4)^{2}+k=-1\Rightarrow a(1)+k=-1 \). Using \( x=-4 \), \( g(-4)=0\Rightarrow k = 0 \). So \( a=-1 \). So \( g(x)=-(x + 4)^{2}=-x^{2}-8x - 16 \). The coefficient of \( x^{2} \) is - 1 (negative), so it opens downward, has a maximum at \( x=-4 \) (the vertex). The x - intercept is at \( x=-4 \) (since \( g(x)=-(x + 4)^{2}=0\Rightarrow x=-4 \)). The y - intercept is at \( x = 0\), \( g(0)=-16 \). Now, option A: maximum at x - intercept (x=-4 is x - intercept, and it's the maximum point). Option B: minimum at y - intercept? No, the function has a maximum, not a minimum in the sense of a vertex - based minimum (it opens downward). Option C: minimum at x - intercept? No, the function has a maximum at x - intercept. Option D: maximum at y - intercept? No, y - intercept is at (0,-16), and the maximum is at (-4,0).

Answer:

A. The maximum occurs at the function's x - intercept.