Question

1. an absolute value equation is shown.

( y = -2|x + 7| - 4 )
which of the following is the graph of the equation?
a.
graph a
b.
graph b
c.
graph c
d.
graph d

1. the table shows selected values for a function ( f(x) = |x - h| + m )

( x )	-2	0	2	4
( f(x) )	6	4	2	4

what are the coordinates of the vertex of the function?
a. ( (2, 0) )
b. ( (2, 2) )
c. ( (0, 4) )
d. ( (-2, 6) )

Explanation:

Question 1 (Graph of Absolute Value Equation)

Step1: Recall Absolute Value Form

The equation is \( y = -2|x + 7| - 4 \). The vertex form of an absolute value function is \( y = a|x - h| + k \), where \((h, k)\) is the vertex. Here, \( h = -7 \), \( k = -4 \), and \( a = -2 \) (negative, so the graph opens downward).

Step2: Analyze Vertex and Direction

The vertex is at \((-7, -4)\). The coefficient \( a = -2 \) means the graph is a vertical stretch (factor 2) and opens downward. Now, check the graphs:

• Graph A: Vertex around \((-7, -4)\)? Let's see the x - axis (negative side) and y - axis (negative). The shape: opens downward, vertex at x = -7 (left side), y = -4 (below zero). This matches the vertex and direction.
• Graph B: Opens upward (since the "V" is upward - shaped), but our \( a=-2\) (downward), so eliminate.
• Graph C: Vertex at positive x, but our vertex is at x = -7 (negative), eliminate.
• Graph D: Opens upward, eliminate.

Step1: Recall Vertex Form of Absolute Value Function

The function is \( f(x)=|x - h|+m \) (wait, maybe a typo, should be \( f(x)=|x - h|+k \)). The vertex of \( y = |x - h|+k \) is at \((h, k)\). For an absolute value function, the vertex occurs where the expression inside the absolute value is zero, i.e., \( x - h = 0\Rightarrow x = h \), and the minimum (or maximum, but here since the coefficient of the absolute value is positive, it's a minimum) value is \( k \).

Step2: Analyze the Table

The table values:

• When \( x=-2 \), \( f(x) = 6 \)
• When \( x = 0 \), \( f(x)=4 \)
• When \( x = 2 \), \( f(x)=2 \)
• When \( x = 4 \), \( f(x)=4 \)

Notice that the function value is minimum at \( x = 2 \) (since \( f(2)=2 \), and then it increases as we move away from \( x = 2 \) (left to \( x=-2 \), right to \( x = 4 \), the values increase: 2→4→6 and 2→4). For the absolute value function \( f(x)=|x - h|+k \), the vertex is at the point where the function reaches its minimum (since the coefficient of the absolute value is positive, so it's a "V" opening upward). The minimum value occurs at \( x = h \), and \( f(h)=k \). Here, the minimum value of \( f(x) \) is 2, which occurs at \( x = 2 \). So the vertex is at \((2, 2)\) (since at \( x = 2 \), \( f(x)=2 \), and this is the minimum point, so \( h = 2 \), \( k = 2 \)).

Let's verify: If \( h = 2 \), then \( f(x)=|x - 2|+k \). When \( x = 2 \), \( f(2)=|0|+k=k \). From the table, \( f(2)=2 \), so \( k = 2 \). So the vertex is \((2, 2)\).

Answer:

A (the graph labeled A)

Question 2 (Vertex of Absolute Value Function)