Question

the graph of the parent quadratic function $f(x)=x^2$ and that of a second function of the form $g(x)=ax^2$ are shown. what conclusion can you make about the value of $a$ in the second function?\
\
a. $a > 1$\
b. $a < -1$\
c. $0 < a < 1$\
d. $-1 < a < 0$

Explanation:

Step1: Recall properties of \( y = ax^2 \)

For the quadratic function \( g(x)=ax^2 \), the sign of \( a \) determines the direction of the parabola (upward if \( a>0 \), downward if \( a<0 \)), and the magnitude of \( |a| \) determines the width (narrower if \( |a|>1 \), wider if \( 0<|a|<1 \)).

Step2: Analyze the graph

The parent function \( f(x)=x^2 \) opens upward (since the coefficient of \( x^2 \) is \( 1>0 \)). The second function \( g(x)=ax^2 \) in the graph opens downward, so \( a < 0 \). Now, compare the width: the graph of \( g(x) \) is narrower (or has a steeper slope) than \( f(x) \) or is a reflection with a vertical stretch? Wait, no—wait, when \( a \) is negative, the parabola opens downward. If we compare the "steepness", for \( |a| > 1 \), the parabola is narrower (steeper), for \( |a| < 1 \), it's wider (flatter). Since the downward - opening parabola \( g(x)=ax^2 \) is narrower (or has a steeper slope) than the upward - opening \( f(x)=x^2 \), the magnitude \( |a|>1 \). Since \( a < 0 \) and \( |a|>1 \), we have \( a < - 1 \).

Answer:

B. \( a < - 1 \)