How do you graph and label the vertex and axis of symmetry of #y=(x+2)^2-3#?

Question

Julia Adams · Accepted Answer

Using the information given to you and a bit of expanding.

Lets start with the known information. From this we can identify the the vertex of the graph which as we know would be #(-2,-3)# The X value is negative as the bracket is equal to 0. We can also work out the roots by solving for x with the given completed square. There set y = to 0 and solve, #(x+2)^2-3 = 0# #therefore (x+2)^2=3# #therefore x+2= +-sqrt(3)# #therefore x =-2+-sqrt(3)# Now we have our roots of the graph. Now finally its time for the y-intercept, this can easily be obtained by expanding out the given completed square. #therefore (x+2)^2-3 = x^2+4x+4-3 = x^2+4x+1# This #implies# that the y intercept #=(0,1)# Now all we need to do is plot the key points and sketch the graph. graph{y=(x+2)^2-3 [-10, 10, -5, 5]}

Eli Assouline · Answer

undefined To graph and label the vertex and axis of symmetry of the equation $ y = (x + 2)^2 - 3 $:

1. The vertex of the parabola is at the point (-2, -3).
2. The axis of symmetry is the vertical line passing through the vertex, which is $ x = -2 $.

You can plot the vertex (-2, -3) on the graph and draw the axis of symmetry as a vertical line at $ x = -2 $.