Question

match each polynomial function to its graph.
$f(x) = x^{4} + 3x^{2} + 2$
$g(x) = -x^{4} - 4x^{2} - 2$

Explanation:

Step1: Analyze the leading term of \( f(x) \)

The leading term of \( f(x)=x^4 + 3x^2+2 \) is \( x^4 \) with a positive coefficient. For even - degree polynomials (\( \text{degree}=4 \)), when the leading coefficient is positive, as \( x
ightarrow\pm\infty \), \( f(x)
ightarrow+\infty \). Also, when \( x = 0 \), \( f(0)=0^4+3\times0^2 + 2=2 \). The graph of a function with a positive leading coefficient for an even degree will open upwards (both ends go to positive infinity) and have a \( y \) - intercept of 2.

Step2: Analyze the leading term of \( g(x) \)

The leading term of \( g(x)=-x^4-4x^2 - 2 \) is \( -x^4 \) with a negative coefficient. For an even - degree polynomial (\( \text{degree} = 4 \)), when the leading coefficient is negative, as \( x
ightarrow\pm\infty \), \( g(x)
ightarrow-\infty \). When \( x = 0 \), \( g(0)=-0^4-4\times0^2-2=-2 \). The graph of a function with a negative leading coefficient for an even degree will open downwards (both ends go to negative infinity) and have a \( y \) - intercept of - 2.

Step3: Match the functions to the graphs

The right - hand graph has a \( y \) - intercept of 2 and opens upwards (both ends go to positive infinity), so it matches \( f(x)=x^4 + 3x^2+2 \). The left - hand graph has a \( y \) - intercept of - 2 (or we can see it opens downwards) and matches \( g(x)=-x^4-4x^2 - 2 \).

Answer:

\( f(x)=x^4 + 3x^2+2 \) matches the right - hand graph; \( g(x)=-x^4-4x^2 - 2 \) matches the left - hand graph.