This week in PreCalc 11, we learned about graphing quadratic inequalities with two variables. Inequalities state whether two values are equal, not equal, greater than or less than the other. The point of the inequality is to make it true, If the inequality states that then it’s true because 7 is greater than 6, but if it states , then it’s not true because 7 isn’t less than 6.

Inequalities can be expressed by quadratic equations. For example . We are supposed to graph this quadratic equation in order to find possible solutions that make the statement true.

Step 1: Graph

To start this process, we have to graph the quadratic. We can start graphing this quadratic by one of two ways. One way is to convert this inequality from general form to vertex form. We can do this by the completing the square method. This form tells us where the parabola is located and what the stretch pattern is so we know the pattern to go up by. Another way we can graph this is to factor it. To factor we have to find two numbers that when multiplied equal -8 and when added equal -2:

In this case the numbers were -4 and 2.

Now we know what the zeros of the equation is so that means the opposites are the x-intercepts. The opposite of -4 and 2 are 4 and -2 which means that those are the x-intercepts of the graph. Since we know where the x-intercepts are, we can find out the vertex. We can do this by adding the x intercepts together and dividing them by 2.

Now we know that 1 is the x value of the vertex. To find the y value, we can put the x value into the original equation to solve for y.

Now we have the y-value which means we know the vertex

We also know that the stretch value, the value that determines the pattern the parabola follows, is 1 because there is no number attached to the in the inequality. That means the parabola follows the pattern of up 1 over 1, up 3 over 1, up 5 over 1, and so on.

The graph should look like the following:

Step 2: Determine the Solution

Now that we know what the parabola looks like, we have to figure out which coordinates on the graph make the inequality statement true. The points that make the statement true are either going to be all the ones inside the parabola or all the ones outside the parabola. To figure this out we can start by picking one point of the graph and dropping the x value and the y value in the original equation. If, when the inequality is solved, the statement is true, then all the points, where its located (i.e. outside the parabola) then all the points outside the parabola are true, so we would shade that area. If it was located inside the parabola then we would shade the inside of the parabola.

When the coordinate is placed in the original inequality, a true statement came out. Zero is greater than negative eight, so that means, since , is inside the parabola, then all of inside the parabola will be shaded.

All the coordinates inside the parabola make the original inequality true.

This week in PreCalc 11, we reviewed and practiced how to read a graph and make an equation in the different forms.

The three forms that we worked with this unit was:

standard form or vertex form- ,where (p, q) is the vertex of the parabola, and a is the stretch value. It’s the easiest equation to graph because we are given a starting point and the pattern that it goes up by.

general form- .  A, B, and C are three real numbers. Once these are given, the values for x and y that make the statement true express a set of (x, y) points which form a parabola when graphed. It’s the most useless equation to graph because we are only given the y-intercept and can’t do much else unless it’s converted to one of the other forms.

factored form– . x1 and x2 are the opposite of the x-intercepts of a graph. If the x intercepts are +3 and +5, then x1 would be -3 and x2 would be -5,  and the a value is the stretch value. This equation is useful only when given the x-intercepts and is used mostly when trying to search for the a value as long as there is another point given along the parabola to substitute y and x for to find a.

It’s simple to convert the equations between each other because it uses all the skills that we have learned and used earlier in the semester.

Example:

Step 1: Turning It Into Vertex Form

First we have to look at the vertex and see how far it has moved from (0,0). In this picture the vertex has moved 3 units to the left and 2 units down. The 3 units left is our p-value and the 2 units down is our q-value. If we put it into the vertex equation equation, we are left with:

Now we have the a-value left to find. If the a-value is one then the pattern the parabola would go up is up 1 over 1, up 3 over 1, up 5 over 1, up 7 over one, etc…Typically the amount the first jump goes up by is usually the a-value. If the first jump is up 2 and over one, then the pattern is probably 2. That means the original pattern of up 1 over 1, up 3 over 1 would be multiplied by 2 so the 2 pattern is up 2 over one, up 6 over one, up 10 over one, etc… In this case the pattern is 2 because the first jump is over up 2 over one so that’s the a-value. That means the complete vertex equation is:

Step 2: Convert Vertex Form into General Form

To convert from vertex form to general form is quite simple since it’s mostly expanding and simplifying.

Now that we have , you just expand and simplify!

Now we have general form of the equation!

Step 3: Turn the Equation into Factored Form

There are 2 ways that we can turn the graph into a factored form equation

First Way:

The first way is to factor the general form:

Step 1 is to factor out the 2 from the equation

Step 2 is to find two numbers that when multiplied equal 3 and when added equal 4. In this case, it’s 3 and 1.

Step 3 is to find a number that when added to or subtracted from +3 and +1 will equal zero. In this case, the answer is -3 and -1. That means we have found the x-intercepts. We also found out the stretch value.

Second Way:

Second way to get it into factored form is to look at the graph and find the x-intercepts. The x-intercept is where the parabola crosses the x axis. There may 2, or one, or even none. In this case, there are 2 and the x-intercepts are -3 and -1.

Since we are given the x-intercepts we can place them directly into the equation

We also know the a-value which is 2 so we can put the a value in. But if we didn’t have the a-value then we could put in another point and substitute x and y for the coordinates, but we know a so we don’t need to do that.

Now we have the factored form, and we know how to convert between the rest of equations!