# How do I graph y= l (x^2) +4x l (absolute values) and please go through the steps. Thanks

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### 1 Answer

You need to use absolute value definition, such that:

`|x^2+4x| = {(x^2 + 4x, x^2 + 4x >= 0),(-x^2 - 4x, x^2 + 4x < 0):}`

You need to solve for x the equation `x^2 + 4x = 0` , such that:

`x^2 + 4x = 0 => x(x + 4) = 0`

Using zero product rule, yields:

`x_1 = 0`

`x + 4 = 0 => x_2 = -4`

Hence the equation of the function `f(x)` is `f(x) = x^2 + 4x` for `x in (-oo,-4]U[0,oo)` and it is `f(x) = -x^2 - 4x` if `x in (-4,0)`

Since the equation `x^2 + 4x = 0` has two roots `x_1 = -4` and `x_2 = 0` , yields that the graph of the function intersects x axis at `x_1 = -4` and `x_2 = 0` .

The graph of the function is a parabola that opens upward, because the leading coefficient a = 1 is positive, hence, the vertex of parabola is a minimum point.

You may evaluate the coordinates of vertex of parabola `f(x) = x^2 + 4x` , using the formula `(-b/(2a),(4ac-b^2)/(4a))`

`(-b/(2a),(4ac-b^2)/(4a)) = (-4/2,-16/4) = (-2,-4)`

The graph of the function `f(x) = -x^2 - 4x` is a parabola that opens downward since `a = -1 < 0` and it has a maximum point at `(2,4)` .

**The graph of the given function `f(x) = |x^2 + 4x|` is sketched below, such that:**