Question

which function has a minimum and is transformed to the right and down from the parent function, $f(x) = x^2$?
\\(\bigcirc\\) $g(x) = -9(x + 1)^2 - 7$
\\(\bigcirc\\) $g(x) = 4(x - 3)^2 + 1$
\\(\bigcirc\\) $g(x) = -3(x - 4)^2 - 6$
\\(\bigcirc\\) $g(x) = 8(x - 3)^2 - 5$

Explanation:

Step1: Recall vertex form properties

The vertex form of a quadratic function is \( g(x)=a(x - h)^2 + k \), where \( (h,k) \) is the vertex, and \( a \) determines the direction (up/down) and width. If \( a>0 \), the parabola opens up (has a minimum); if \( a<0 \), it opens down (has a maximum). A horizontal shift right is \( h>0 \) (since it's \( x - h \)), and a vertical shift down is \( k<0 \).

Step2: Analyze each option

• Option 1: \( g(x)=-9(x + 1)^2 - 7 \)

Rewrite \( x + 1 \) as \( x - (-1) \), so \( h=-1 \) (shift left), \( a=-9<0 \) (opens down, has maximum). Eliminate.

• Option 2: \( g(x)=4(x - 3)^2 + 1 \)

\( a = 4>0 \) (opens up, minimum), \( h = 3>0 \) (shift right), but \( k = 1>0 \) (shift up). We need shift down, so eliminate.

• Option 3: \( g(x)=-3(x - 4)^2 - 6 \)

\( a=-3<0 \) (opens down, maximum). Eliminate.

• Option 4: \( g(x)=8(x - 3)^2 - 5 \)

\( a = 8>0 \) (opens up, minimum), \( h = 3>0 \) (shift right), \( k=-5<0 \) (shift down). This matches all criteria.

Answer:

\( \boldsymbol{g(x) = 8(x - 3)^2 - 5} \) (the fourth option)