Question

graph $h(x) = 8|x + 1| - 1$.

Explanation:

Step1: Recall the vertex form of absolute value function

The general form of an absolute value function is \( y = a|x - h| + k \), where \((h, k)\) is the vertex of the V - shaped graph. For the function \( h(x)=8|x + 1|-1 \), we can rewrite \( x + 1\) as \( x-(-1) \), so comparing with \( y=a|x - h|+k \), we have \( h=-1 \), \( k = - 1\) and \( a = 8\). So the vertex of the graph should be at the point \((-1,-1)\). But in the given graph, the vertex is at \((0,0)\), which means there is a mistake in the graphing of the function \( h(x)=8|x + 1|-1\).

Step2: Analyze the slope and transformation

The coefficient \( a = 8\) determines the slope of the two branches of the absolute - value graph. For \( x\geq - 1\), the function \( h(x)=8(x + 1)-1=8x+8 - 1=8x + 7\). The slope of this line is \( 8\), which is a very steep positive slope. For \( x\lt - 1\), the function \( h(x)=-8(x + 1)-1=-8x-8 - 1=-8x - 9\), with a slope of \(-8\), a very steep negative slope.

Let's find a point on the correct graph. When \( x = 0\), \( h(0)=8|0 + 1|-1=8\times1-1 = 7\). So the point \((0,7)\) should be on the graph. When \( x = 1\), \( h(1)=8|1 + 1|-1=8\times2-1 = 15\). The given graph has a much shallower slope and an incorrect vertex, so it does not represent the function \( h(x)=8|x + 1|-1\) correctly.

Answer:

The given graph does not correctly represent the function \( h(x)=8|x + 1|-1\). The correct vertex should be at \((-1,-1)\), and the slopes of the two branches should be \(8\) (for \(x\geq - 1\)) and \(-8\) (for \(x\lt - 1\)), leading to a steeper graph with a vertex at \((-1,-1)\) rather than \((0,0)\) as shown.