Question

How do you graph the function `y=1/2x^2+2x-3/8` and identify the domain and range?

Answer 1

Domain: `(-oo, oo)`
Range: `(1/2, oo)`

Explanation:

How to graph the function:

1) Find the zeros or roots of the function

This can be done by factoring the quadratic equation.

In this case your zeros are: `-2+-sqrt(4.75)`.

This can be found using the quadratic formula: `(-b+-sqrt(b^2-4ac))/(2a)`. Where `a`, `b`, and `c` are the coefficients of the terms in your quadratic equation.

2) Find the vertex

This can be done by rewriting your quadratic equation into vertex form: `y = a*(x-h)^2 + k`
Where `h` and `k` are the `x` and `y` coordinates of the vertex.

3) You can insert more values into `x` and calculate the `y` coordinate. This step is optional because you should be able to draw the parabola with the 3 points already discovered.

How to identify the domain and range

1) Domain

The domain of any quadratic is always `(-oo, oo)` this is because no matter what `x` value you choose from `-oo` to `oo` there is always a `y`.

2) Range

The range of any quadratic is from the vertex either all values below or above.

To know whether the parabola faces up like a U or down like an upside down U you look for the sign of the leading coefficient, or `a`. In this case the leading coefficient is `1/2`, a positive number. Therefore in this case the range consists of the vertex point and all values above it. So the range is `(1/2, oo)`