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?\
\
\bigcirc a. $a < -1$\
\bigcirc b. $0 < a < 1$\
\bigcirc c. $-1 < a < 0$\
\bigcirc d. $a > 1$

Explanation:

Step1: Analyze the direction of the parabola

The parent function \( f(x) = x^2 \) opens upwards (since the coefficient of \( x^2 \) is positive). The second function \( g(x)=ax^2 \) in the graph opens downwards, so the coefficient \( a \) must be negative (because if \( a>0 \), the parabola opens up; if \( a<0 \), it opens down). So we can eliminate options B (\( 0 < a < 1 \)) and D (\( a > 1 \)) as they imply \( a \) is positive.

Step2: Analyze the vertical stretch/compression

For a quadratic function \( y = ax^2 \), the magnitude of \( |a| \) determines the vertical stretch or compression. If \( |a|>1 \), the parabola is narrower than \( y = x^2 \); if \( 0<|a|<1 \), it is wider. Looking at the graph, the downward - opening parabola ( \( g(x) = ax^2 \)) is narrower than the upward - opening parent parabola \( f(x)=x^2 \)? Wait, no. Wait, the parent function \( f(x)=x^2 \) and the other parabola: when \( a \) is negative, let's compare the width. The standard parabola \( y = x^2 \) and \( y=ax^2 \) with \( a<0 \). If \( |a|>1 \), the parabola \( y = ax^2 \) is narrower. If \( |a|<1 \), it is wider. Looking at the graph, the downward - opening parabola (the one with \( a<0 \)) is narrower than the upward - opening \( y = x^2 \)? Wait, no, actually, in the graph, the two parabolas: the upper one ( \( y = x^2 \)) and the lower one ( \( y=ax^2 \)). Wait, no, the parent function \( f(x)=x^2 \) is the upper (opening up) parabola. The other parabola (opening down) has a "narrower" or "wider" shape? Wait, when \( a=- 2 \), the parabola \( y=-2x^2 \) is narrower than \( y = x^2 \), when \( a = - 0.5 \), it is wider. Looking at the graph, the downward - opening parabola is narrower than the upward - opening \( y=x^2 \). So \( |a|>1 \). Since \( a<0 \), then \( a < - 1 \). So option A (\( a < - 1 \)) is correct, and option C (\( - 1 < a < 0 \)) would imply that \( |a|<1 \), which would mean the downward - opening parabola is wider than \( y = x^2 \), which is not the case here.

Answer:

A. \( a < - 1 \)