Question

What would be the interval of decrease of this quadratic function? f(x)=x²

Answer 1

`-oo < x < 0`

Explanation:

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`f(x)=x^2` is the equation of a parabola. In calculus, there are specific methods for determining such intervals using derivatives of functions.

But since this problem is posted as an algebra problem, I will assume that the student has not had calculus yet. As such, we will approach this differently.

The coefficient of `x^2` is `+1`. A positive coefficient indicates that the parabola opens up. This means that the vertex of the parabola is where the function has its minimum.

As such, the function decreases between `-oo` and the `x`-coordinate of the vertex; and it increases between that point and `+oo`.

Let's figure out the coordinates of the vertex. If the equation of the function is in the form of:

`f(x)=y=ax^2+bx+c`

Then the `x`-coordinate of the vertex can be found using the following formula:

`x_(vertex)=-b/(2a)`

In our equation, `a=1, b=0, and c=0`.

`x_(vertex)=-0/(2(1))=-0/2=0`

The `y`-coordinate of the vertex can be found by plugging this `x` value into the equation:

`y_(vertex)=(0)^2=0`

`Vertex(0,0)`

Interval of decrease is:

`-oo < x < 0`

You can see this in the graph of the function below:

graph{x^2 [-10, 10, -5, 5]}