Question

match each quadratic function to its graph.
$f(x) = x^2 + 4$
$g(x) = x^2 + 12x + 32 = (x + 8)(x + 4)$
(graphs are shown below the functions, with two grid - based graphs of parabolas and another curve, and two blue boxes with the functions $f(x)=x^2 + 4$ and $g(x)=x^2 + 12x + 32$ above the graphs)

Explanation:

Step1: Analyze \( f(x) = x^2 + 4 \)

The function \( f(x)=x^2 + 4 \) is a quadratic in vertex form (or standard form with vertex at \( (0, 4) \)) since it can be written as \( f(x)=(x - 0)^2+4 \). The vertex is at \( (0, 4) \), and it opens upwards (because the coefficient of \( x^2 \) is positive). So we look for a parabola with vertex on the y - axis (x = 0) at y = 4.

Step2: Analyze \( g(x)=x^2 + 12x + 32=(x + 8)(x + 4) \)

First, find the roots by setting \( g(x)=0 \). If \( (x + 8)(x + 4)=0 \), then \( x=-8 \) or \( x = - 4 \). The vertex of a quadratic \( ax^2+bx + c \) is at \( x=-\frac{b}{2a} \). For \( g(x)=x^2+12x + 32 \), \( a = 1 \), \( b = 12 \), so \( x=-\frac{12}{2\times1}=-6 \). Then \( g(-6)=(-6)^2+12\times(-6)+32=36-72 + 32=-4 \). So the vertex is at \( (-6,-4) \), and it opens upwards. We look for a parabola with roots at \( x=-8 \) and \( x=-4 \) (crossing the x - axis at these points) and vertex at \( (-6,-4) \).

Step3: Match the graphs

• For \( f(x)=x^2 + 4 \): The right - hand graph (the one with the orange parabola having vertex at (0,4)) matches because its vertex is on the y - axis at y = 4.
• For \( g(x)=x^2+12x + 32 \): The left - hand graph (the orange parabola crossing the x - axis at \( x=-8 \) and \( x=-4 \) and with vertex at \( (-6,-4) \)) matches.

Answer:

• \( f(x)=x^2 + 4 \) matches the right - hand graph (with vertex at (0,4)).
• \( g(x)=x^2 + 12x + 32 \) matches the left - hand graph (with roots at \( x=-8 \) and \( x=-4 \) and vertex at \( (-6,-4) \)).