Given #y=-2(x+4)^2-3#, how do you find the vertex and axis of symmetry?

1 Answer
Jan 9, 2017

#"vertex " =(-4,-3)," axis of symmetry is " x=-4#

Explanation:

The equation of a parabola in #color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#
where (h ,k) are the coordinates of the vertex.

#y=-2(x+4)^2-3" is in this form"#

#" Note " (x+4)^2=(x-(-4))^2#

#"by comparison " h=-4" and " k=-3#

#rArr" vertex " =(-4 ,-3)#

The axis of symmetry passes through the vertex.

Since the coefficient of #(x+4)^2" is negative, then the parabola opens down"#

#rArr"equation of vertex is " x=-4#
graph{-2(x+4)^2-3 [-10, 10, -5, 5]}