Question

match each quadratic function to its graph.
$f(x) = 2x^2 + 28x + 98 = 2(x + 7)^2$
$g(x) = -x^2 - 7x - 10 = -(x + 5)(x + 2)$
$h(x) = x^2 + 3$
$k(x) = -x^2 - 4$
$f(x) = 2x^2 + 28x + 98$
$g(x) = -x^2 - 7x - 10$
$h(x) = x^2 + 3$
$k(x) = -x^2 - 4$

Explanation:

Step1: Analyze \( f(x) = 2(x + 7)^2 \)

The vertex form of a quadratic is \( a(x - h)^2 + k \), where \((h, k)\) is the vertex. For \( f(x) \), \( h = -7 \), \( k = 0 \), and \( a = 2>0 \) (opens upward). So the graph with vertex at \( (-7, 0) \) and opening up matches \( f(x) \) (top - left graph).

Step2: Analyze \( g(x)=-(x + 5)(x + 2) \)

First, find the roots: set \( g(x)=0 \), then \( x=-5 \) or \( x=-2 \). The coefficient of \( x^2 \) is \(-1<0\) (opens downward). The graph with roots at \(-5\) and \(-2\) and opening down (bottom - right graph) matches \( g(x) \).

Step3: Analyze \( h(x)=x^2 + 3 \)

Vertex form: \( h(x)=(x - 0)^2+3 \), vertex at \((0, 3)\), \( a = 1>0 \) (opens upward). The graph with vertex at \((0, 3)\) and opening up (bottom - left graph) matches \( h(x) \).

Step4: Analyze \( k(x)=-x^2-4 \)

Vertex form: \( k(x)=-(x - 0)^2-4 \), vertex at \((0, - 4)\), \( a=-1<0\) (opens downward). The graph with vertex at \((0, - 4)\) and opening down (top - right graph) matches \( k(x) \).

Answer:

• \( f(x)=2x^{2}+28x + 98 \): Top - Left Graph
• \( g(x)=-x^{2}-7x - 10 \): Bottom - Right Graph
• \( h(x)=x^{2}+3 \): Bottom - Left Graph
• \( k(x)=-x^{2}-4 \): Top - Right Graph