Quadratics come in two forms:

#f(x)=ax^2+bx+c# #color(blue)(" Standard Form")#

#f(x)=a(x-h)^2+k# #color(blue)(" Vertex Form")#

Obviously we'll ignore the #"standard form"# for this problem, but it's important to know both.

Since our equation is in #"vertex"# form, we're given the #"vertex"# without having to solve for it:

#"Vertex: " (-h, k)#

Don't forget that the default vertex is #-h#, don't forget the negative! Let's look back at our original equation:

#f(x)=(xcolor(red)(+7))^2color(red)(""-5)#

Let's plug in our #h# and #k# values into the #"vertex point:"#

#(-h, k)#

#((-)+7, -5)#

#color(red)((-7, -5)#

Notice that a negative and a positive make a negative, hence the #-7# even though it's #+7# in the original equation.

Now that we know our #"vertex"#, solving for domain and range is very easy

#"Domain: All x-values"#

The good thing with this problem is that all quadratics will always have an infinite domain of #" all real numbers "# since the graph goes on infinite horizontally and vertically (upward). So:

#color(red)(D: (-oo, oo))#

#"Range: All y-values"#

With both #"Domain"# and #"Range"# we measure from lowest to highest, and the lowest point on this quadratic is the #y"-coordinate"# of the vertex since the graph opens upward infinitely. So:

#color(red)(R: [-5, oo))#

It's important to remember that when a value is included or "touched" on the graph for domain and/or range, it must have a bracket. If it has a parentheses, it means that it goes up to that value, but does not touch it, like an asymptote. Obviously, we can't touch infinity, so we leave those as parentheses, but the graph does touch -5, so we use brackets on that part, but not the infinity.

To better understand what these answers mean, it's better to read them in a sentence:

The #"Domain"# reads as #"the graph includes all x-values,"# since the quadratic does not end horizontally.

The #"Range"# reads as #"The graph starts at "-5" and extends upward infinitely."#

If you still are confused you can always visualize it:

graph{(x+7)^2-5 [-10, 10, -5, 5]}