What is the the vertex of #y = -x^2 - 3#?

2 Answers
Jan 5, 2016

#Vertex:(0,-3)#

Explanation:

#y=-x^2-3#
Let us first convert this in vertex from

#color(brown)"vertex form:y=a(x-h)^2+k"#
#color(brown)"vetex:(h,k)"#
Let us write the given equation in vertex form.
#y=(x-0)^2+(-3)#
#Vertex:(0,-3)#

Jan 5, 2016

#"vertex"-> (x,y) -> (0,-3)#

Explanation shows what is happening.

Explanation:

Suppose we hade the general equation of #y_1=-x^2#

Then the graph would look like:
Tony B

Subtract 3 from both sides of the equation. Not only is the equation now #y_1 - 3 = -x^3 - 3# but you have lowered the whole thing by 3.
Let #y_1-3# be written as #y_2# now giving: #y_2=x^2-3#
This graph looks like:
Tony B

From this you can see that the vertex in the #color(blue)("first case")# is at #x_("vertex")=0" and "y_("vertex")=0# written as #"vertex" ->(x,y) -> (0,0)#

In the #color(blue)("second case")# it has lowered by 3 on the x-axis giving #x_("vertex")=0" and "y_("vertex")=-3# written as

#"vertex"-> (x,y) -> (0,-3)#