Question

1. which of the following functions has a graph with a vertex that is a translation 6 units horizontally to the left of the vertex of the graph of $f(x) = (x - 4)^2 + 8$?

a $h(x) = (x - 4)^2 + 14$
b $h(x) = (x + 2)^2 + 8$
c $h(x) = (x - 10)^2 + 8$
d $h(x) = (x - 4)^2 + 2$

Explanation:

Step1: Find vertex of \( f(x) \)

The vertex form of a parabola is \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. For \( f(x) = (x - 4)^2 + 8 \), the vertex is \((4, 8)\).

Step2: Determine horizontal translation

A horizontal translation 6 units left means the new \( h \)-value is \( 4 - 6 = -2 \). So the vertex form should have \( (x - (-2))^2 = (x + 2)^2 \), with the same \( k \)-value (8) for vertical position.

Step3: Match with options

Check each option:

• Option A: \( h(x) = (x - 4)^2 + 14 \) (vertical translation, not horizontal).
• Option B: \( h(x) = (x + 2)^2 + 8 \) (vertex \((-2, 8)\), which is 6 left of \((4, 8)\)).
• Option C: \( h(x) = (x - 10)^2 + 8 \) (translation 6 right).
• Option D: \( h(x) = (x - 4)^2 + 2 \) (vertical translation, not horizontal).

Answer:

B. \( h(x) = (x + 2)^2 + 8 \)