Question

match each quadratic function to its graph.\\(f(x) = x^2 - 14x + 51\\)\\(g(x) = -x^2 - 4\\)

Explanation:

Step1: Analyze \( f(x) = x^2 - 14x + 51 \)

The coefficient of \( x^2 \) is positive (\( 1>0 \)), so the parabola opens upward. To find the vertex, we complete the square or use the formula \( x = -\frac{b}{2a} \). For \( f(x) \), \( a = 1 \), \( b = -14 \), so \( x = -\frac{-14}{2\times1} = 7 \). Substitute \( x = 7 \) into \( f(x) \): \( f(7) = 7^2 - 14\times7 + 51 = 49 - 98 + 51 = 2 \). So the vertex is \( (7, 2) \), which matches the right - hand graph (opening upward, vertex at \( x = 7 \), \( y = 2 \)).

Step2: Analyze \( g(x)=-x^{2}-4 \)

The coefficient of \( x^2 \) is negative (\( - 1<0 \)), so the parabola opens downward. The vertex form of a parabola is \( y = a(x - h)^2 + k \), and for \( g(x)=-x^{2}-4 \), it can be written as \( g(x)=-(x - 0)^2-4 \), so the vertex is \( (0,-4) \), which matches the left - hand graph (opening downward, vertex at \( (0, - 4) \)).

Answer:

\( f(x)=x^{2}-14x + 51 \) matches the right - hand graph (with vertex \((7,2)\) and opening upward), \( g(x)=-x^{2}-4 \) matches the left - hand graph (with vertex \((0, - 4)\) and opening downward).