# Week 9 – precalculus 11

Equivalent Forms of the Equation of a Quadratic Functions and finding x-ints:

If one is given the equation $y=2x^2 +12x+24$ it may seem like there is no way to find the vertex or starting point of the parabola, a line that goes down but then goes back up after a certain point. If we needed to find the x-intercepts we would factor this equation. If we wanted the vertex we would complete the square. The equation we are given is in general form but we can change it to vertex form/standard form. We can do this the other way around as well. For this example we will find the standard form and factored form.

Standard form: $y=a(x-p)^2 +q$

$y=2x^2 +12x+24$

We will take out the two and start the process of completing the squar

$y=2(x^2 +6x)+24$

Then expand out and make a zero pair

$y=2(x^2 +6x +9 -9)+24$

Then take out the -9

$y=2(x^2 +6x +9)+24-18$

And factor the brackets

$y=2(x+3)^2+6$

And now we have the vertex and the stretch value. The parabola will be skinnier because of the strech of 2 and it will be 3 left and 6 up. I explained these values in my earlier blog post but the main idea is that the number outside in front of the brackets is the stretch and the one inside is the x axis and determines which way it goes (-is right +is left) and the number we add at the end is the placement on the y- axis. The numbers determine where the vertex of the parabola is.

If we wanted to find the x-intercepts we would factor the general form. We will use a new equation this time

$y=2x^2 -18x+40$

Take out the 2

$y=2(x^2 -9x+20)$

The we factor, -4 times -5 is 20 but adding them together gives us -9, which is perfect.

$y=2(x-5)(x-4)$

The factored form takes the format of $y= a (x-x_1)(x-x_2)$ The numbers in the x1 and x2 place will give us the x-ints. So in this case the x-ints are 5 and 4 because we are subtracting +5 and +4

Just by changing the form of an equation can we find new information about the parabola it represents. All of these equations are equal.

To get back to General form from Standard form we just expand.

$y=2(x+3)^2+6$

Then

$y=2(x+3)(x+3)+6$ $y=2(x^2 +6x +9) +6$

then

$y=2x^2+ 12x +18 +6$

and

$y=2x^2+ 12x +24$

It is now identical to the first equation we used. Now we can find the x-ints if given the Standard form by converting to General form first. We could also use this to check our work and make sure we completed the square or factored correctly.