Question

question 1 of 10
which describes the graph of $y = (x + 6)^2 + 1$?

a. vertex at $(-6, -1)$
b. vertex at $(-6, 1)$
c. vertex at $(6, -1)$
d. vertex at $(6, 1)$

Explanation:

Step1: Recall vertex form of parabola

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

Step2: Compare with given equation

Given equation is \( y=(x + 6)^2+1 \). We can rewrite \( x + 6 \) as \( x-(-6) \). So comparing with \( y = a(x - h)^2 + k \), we have \( h=-6 \) and \( k = 1 \). So the vertex is \((-6,1)\).

Answer:

B. Vertex at \((-6, 1)\)