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Question
identify the vertex, axis of symmetry, and direction of the graph of each function. compare the width of the graph to the width of the graph of ( f(x)=x^2 ) ( h(x)=-8(x + 1)^2 - 1 ) the graph opens down. compare the width of the graph to the width of the graph of ( f(x)=x^2 ) \\(\bigcirc\\) a. the graph is wider than ( f(x)=x^2 ). \\(\bigcirc\\) b. the graph has the same width as ( f(x)=x^2 ). \\(\bigcirc\\) c. the graph is narrower than ( f(x)=x^2 ).
Step1: Recall the vertex form of a parabola
The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex and \(a\) determines the direction and width. For \(f(x)=x^2\), \(a = 1\). For \(h(x)=-8(x + 1)^2 - 1\), we can rewrite \(x + 1\) as \(x - (-1)\), so the vertex is \((-1, -1)\), and \(a=-8\).
Step2: Analyze the width based on \(|a|\)
The width of a parabola \(y = a(x - h)^2 + k\) is determined by \(|a|\). If \(|a|>1\), the parabola is narrower than \(y = x^2\) (where \(|a| = 1\)); if \(|a| = 1\), it has the same width; if \(0<|a|<1\), it is wider. For \(h(x)\), \(|a|=|-8| = 8\), and \(8>1\). So the graph of \(h(x)\) is narrower than \(f(x)=x^2\).
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C. The graph is narrower than \(f(x)=x^2\).