Given function : y = (x+1)^2 State the domain and rangePlease do it in steps, so that its easy for me to understand. Thanks

If possible, you need to put the formula into the
form:

y = ax^2 + bx + c

Once
you've done that, you know the domain is the set of all real
numbers.

For this equation, square (x+1) and
simplify:

y = (x+1)^2 - 8

y =
(x+1)(x+1) - 8

y = x^2 + 1x + 1x + 1 -
8

y = x^2 + 2x - 7

Now,
determine whether the parabola opens up or down. If the a value is positive, it opens
up. In this case, x^2 is the same as (+1)x^2, so a is +1, which is positive, and the
parabola opens upward.

If the parabola opens upward, the
range will be [y-coord. of vertex, infinity). The y-coord. of the vertex is c, which in
this case is -7.

So, the domain is all real numbers, and
the range is [-7, `oo`).

Hope this helps. Good
luck!