Question

for $g(x) = (x + 3)^4$, what happens to the graph at the x - intercept?
options:

• oscillates near $x = -3$
• has a hole at $x = -3$
• touches and turns at $x = -3$
• crosses straight through $x = -3$
• has a vertical asymptote at $-3$

Explanation:

Step1: Analyze the function type

The function \( g(x)=(x + 3)^4 \) is a polynomial function. To find the \( x \)-intercept, we set \( g(x)=0 \), so \( (x + 3)^4=0 \), which gives \( x=-3 \) (a root with multiplicity 4).

Step2: Recall the behavior of polynomial graphs at roots

For a polynomial \( f(x)=(x - a)^n \):

• If \( n \) is even, the graph touches the \( x \)-axis at \( x = a \) and turns around (since the sign of the function doesn't change as \( x \) crosses \( a \), and the graph "bounces" off the axis).
• If \( n \) is odd, the graph crosses the \( x \)-axis at \( x = a \).

Here, \( a=-3 \) and \( n = 4 \) (even). So the graph of \( g(x) \) at \( x=-3 \) will touch the \( x \)-axis and turn around.

Answer:

Touches and turns at \( x=-3 \) (the option with "Touches and turns at \( x=-3 \)")