Given function : y = (x+1)^2 - 8 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!