This week we spoke about inequalities in parabolas and how to solve them algebraically.

Some of these equations may look like x² + 7x + 10 ≥ 0

In these equations, you would first want to solve and factor the equation in order to find the x-intercepts for the parabola being used. The equation also differs if the parabola is a maximum or a minimum. For example, we will say this example is using a minimum parabola, meaning it is opening upwards.

You would work through the equation like this:

Now that you know the x-intercepts of the parabola and that it is a minimum, you need to look at the sign in the equation and which way it is facing. In this equation the sign is facing to the left side, showing that we are looking for the area of the parabola where the lines are more or equal to zero. So, if we are focusing on the lines where they are more or equal to zero then we need to look at the outside sections of the parabola.

This is what the outside sections look like on a parabola:

If the middle/ inside area of the parabola is below zero, then we know that (x≤-5, x≥-2) is the right solution because it is showing the lines above zero.

If a question like x²-36≤0 comes up, it is asking you to find the area where the parabola is less than zero. Again you would have to factor just like this:

Once you find the x-intercept being 6,-6, then you have to represent that to make the answer true and represent the inside on the parabola and where the line is under zero.

looking at this, the answer would be -6≤x≤6 because it represents the area inside the parabola where the line is below zero.