This is a parabola of the form #y = a(x- p)^2 + q#, also known as vertex form. In this form, the vertex is at #(p, q)#.
So, the vertex is at #(-5, -6)#.
Since parameter #a# is positive, the graph opens upwards. Therefore, the graph will have x-intercepts, which are points where #y = 0# on the function, where it touches the x-axis. You can find the x-intercepts by setting #f(x) = 0# and solving.
#0 = (x + 5)^2 - 6#
#6 = (x + 5)^2#
#+-sqrt(6) = x + 5#
#x = -5 +- sqrt(6)#
Some people like approximations for such irrational numbers to make them easier to place precisely on the graph. However, one could say that since #√6# is a little less than halfway between #√4# and #√9#, a fair approximation would be #2..#. The actual value of #sqrt(6)# is #2.4494897...#. We can say using our rough approximation that the x-intercepts are at #(-7.4, 0)# and #(-2.6, 0)#.
The next step towards graphing would be finding the y-intercept. This is when the function touches the y-axis, which is when #x = 0#. This point can be found by evaluating #f(0)#.
#f(0) = (0 + 5)^2 - 6 = 25 - 6 = 19#
#:.# The y-intercept is located at #(0, 19)#.
You can finally perhaps use a table of values to find a few more points on the function--it's totally up to you.
Here's the graph of the function.
graph{y = (x + 5)^2 - 6 [-20, 20, -10, 10]}
Hopefully this helps!
