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- y = a (x-p)^2 + q,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- y = ax^2 + bx + c.  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 formy = a(x-x1)(x-x2). 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:

y = a(x-p)^2 +q

 

y = a (x-(-2)^2 + (-2)

 

y = a (x+2)^2 - 2

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:

y = 2 (x+2)^2 - 2

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 y = 2 (x+2)^2 - 2, you just expand and simplify!

y = 2 (x+2)^2 - 2

 

y = 2 (x+2)(x+2) -2

 

y = 2 (x^2 + 4x +4) -2

 

y = (2x^2+ 8x +6)

 

y = 2x^2 + 8x +6

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

y = 2x^2 + 8x +6

 

y = 2 (x^2 + 4x + 3)

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.

y = 2 (x + 3)(x + 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

y = a (x - x1)(x - x2)

 

y = a (x- (-3)(x - (-1)

 

y = a (x + 3)(x + 1)

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.

y = 2 (x + 3)(x + 1)

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