Question

match each quadratic function to its graph.$f(x) = -x^2 - 13x - 42 = -(x + 7)(x + 6)$$g(x) = x^2 + 3x + 2 = (x + 2)(x + 1)$$h(x) = -x^2 - 1$$k(x) = -2x^2 + 32x - 128 = -2(x - 8)^2$$f(x) = -x^2 - 13x - 42$$g(x) = x^2 + 3x + 2$$h(x) = -x^2 - 1$$k(x) = -2x^2 + 32x - 128$

Explanation:

Step1: Analyze $f(x)=-x^2-13x-42$

Find roots: $-(x+7)(x+6)=0 \implies x=-7, x=-6$. The parabola opens downward (negative leading coefficient), vertex at $x=-\frac{b}{2a}=-\frac{-13}{2(-1)}=-6.5$, which is between -7 and -6. This matches the top-left graph.

Step2: Analyze $g(x)=x^2+3x+2$

Find roots: $(x+2)(x+1)=0 \implies x=-2, x=-1$. The parabola opens upward (positive leading coefficient), vertex at $x=-\frac{3}{2(1)}=-1.5$, between -2 and -1. This matches the bottom-right graph.

Step3: Analyze $h(x)=-x^2-1$

No real roots (set $-x^2-1=0 \implies x^2=-1$). The parabola opens downward, vertex at $(0, -1)$. This matches the bottom-left graph.

Step4: Analyze $k(x)=-2x^2+32x-128$

Rewrite as $-2(x-8)^2$, so it has a repeated root at $x=8$. The parabola opens downward, vertex at $(8, 0)$. This matches the top-right graph.

Answer:

• Top-left graph: $f(x) = -x^2 - 13x - 42$
• Top-right graph: $k(x) = -2x^2 + 32x - 128$
• Bottom-left graph: $h(x) = -x^2 - 1$
• Bottom-right graph: $g(x) = x^2 + 3x + 2$