Question

match each quadratic function to its graph.
$f(x) = 2x^2 + 16x + 24 = 2(x + 6)(x + 2)$
$g(x) = -x^2 - 2$
$h(x) = -2x^2 - 12x - 18 = -2(x + 3)^2$
$k(x) = -x^2 - 5x = -x(x + 5)$
$f(x) = 2x^2 + 16x + 24$ $g(x) = -x^2 - 2$
$h(x) = -2x^2 - 12x - 18$ $k(x) = -x^2 - 5x$

Explanation:

for \( f(x) = 2x^2 + 16x + 24 \)

Step 1: Determine the direction of the parabola

The coefficient of \( x^2 \) is \( 2 \), which is positive. So, the parabola opens upwards.

Step 2: Find the x - intercepts

We have \( f(x)=2(x + 6)(x + 2) \). Set \( f(x)=0 \), then \( 2(x + 6)(x + 2)=0 \). This gives \( x=-6 \) or \( x=-2 \). So the x - intercepts are at \( x=-6 \) and \( x=-2 \).

Step 3: Match with the graph

Among the given graphs, the parabola that opens upwards and has x - intercepts at \( x=-6 \) and \( x=-2 \) will be the graph for \( f(x) \). The right - hand side graph (the one with the upward - opening parabola) has x - intercepts around \( x=-6 \) and \( x=-2 \), so \( f(x) \) matches the right - hand graph.

for \( g(x)=-x^2 - 2 \)

Step 1: Determine the direction of the parabola

The coefficient of \( x^2 \) is \( - 1 \), which is negative. So, the parabola opens downwards.

Step 2: Find the y - intercept and vertex

The y - intercept occurs when \( x = 0 \). Substituting \( x = 0 \) into \( g(x) \), we get \( g(0)=-0^2-2=-2 \). The vertex form of a parabola is \( y = a(x - h)^2+k \), for \( g(x)=-x^2-2 \), we can rewrite it as \( g(x)=-(x - 0)^2-2 \), so the vertex is at \( (0,-2) \) and there are no x - intercepts (since \( -x^2-2 = 0\Rightarrow x^2=-2 \), which has no real solutions).

Step 3: Match with the graph

The parabola that opens downwards, has a y - intercept at \( y=-2 \) and no x - intercepts will be the graph for \( g(x) \). The left - hand side graph (the one with the downward - opening parabola) with vertex at \( (0,-2) \) and no x - intercepts matches \( g(x) \).

for \( h(x)=-2x^2-12x - 18 \)

Step 1: Determine the direction of the parabola

The coefficient of \( x^2 \) is \( - 2 \), which is negative. So, the parabola opens downwards.

Step 2: Find the vertex

We have \( h(x)=-2(x + 3)^2 \). The vertex form of a parabola is \( y=a(x - h)^2+k \), here \( h=-3 \) and \( k = 0 \). So the vertex is at \( (-3,0) \), and it is a repeated root (since it is a perfect square), so the parabola touches the x - axis at \( x=-3 \).

Step 3: Match with the graph

The parabola that opens downwards and touches the x - axis at \( x=-3 \) will be the graph for \( h(x) \). Among the downward - opening parabolas, the one that touches the x - axis at \( x=-3 \) (the left - hand graph's parabola with vertex at \( x=-3 \) on the x - axis) matches \( h(x) \).

for \( k(x)=-x^2-5x \)

Step 1: Determine the direction of the parabola

The coefficient of \( x^2 \) is \( - 1 \), which is negative. So, the parabola opens downwards.

Step 2: Find the x - intercepts

We have \( k(x)=-x(x + 5) \). Set \( k(x)=0 \), then \( -x(x + 5)=0 \). This gives \( x = 0 \) or \( x=-5 \). So the x - intercepts are at \( x = 0 \) and \( x=-5 \).

Step 3: Match with the graph

The downward - opening parabola with x - intercepts at \( x = 0 \) and \( x=-5 \) (the left - hand graph's other parabola) matches \( k(x) \).

Answer:

• \( f(x)=2x^2 + 16x + 24 \) matches the right - hand (upward - opening) graph.
• \( g(x)=-x^2 - 2 \) matches the left - hand graph's parabola with vertex at \( (0,-2) \) (the one without x - intercepts).
• \( h(x)=-2x^2-12x - 18 \) matches the left - hand graph's parabola with vertex at \( x=-3 \) (touching the x - axis).
• \( k(x)=-x^2-5x \) matches the left - hand graph's parabola with x - intercepts at \( x = 0 \) and \( x=-5 \).