Quadratic Functions

First off if you ever get an equation with y=x^2 it is a parabola. Ok so, you should finish solving your parabolic equation by completing the square so,

Factor out a -1
y= -(x^2-12x)

Complete the square inside the brackets
y= -(x^2-12x+36)

Because you added a new 36 to the right side, you have to also add it to the left to balance the equation but be careful because there is a negative sign in front of your number so you're adding in a negative number.
y-36= -(x^2 -12x+36)

Now factor your quadratic.
y-36= -(x-6)^2

Isolate your y value and you've got your equation.
y= -(x-6)^2 +36

To algebraically solve for y=0 just substitute 0 for your y value and solve the equation.
To solve it graphically just plot the parabola on a graph and find the point where it crosses the x-axis or the point where y=0.

MathPong:

Quadratic Functions

I need help with a question i got on my math book. It goes like this Solve the equation y = 0 graphically and algebrally. And i got a formula for this and it goes like: y = x*h and the h in the formula is 12 - x so i did this x(12-x) and i got this 12x-x^2. But its very confusing please help me its a Quadratic Function..

Thanks Walt, I just want to talk about this question because there is an easier way to do it.
----------------------------------------------------------------------------------------------------------------------------------------------------

To solve algebraically you have x(12-x)=0
now if you multiply 2 things together and get 0 for an answer it means that one or the other or both must equal 0. Just think about that fo a moment and convince yourself that it is true.
Now
If x(12-x)=0 then
either x=0 or 12-x=0 and if 12-x=0 then x=12
so the algebraic solution is x=0 and x=12
------------------------------------------------------------------------------------------
It is a bit strange to be asked to solve this graphically but because you effectively use the algebraic solution to graph it anyway.
But i will talk about it.
y= -x^2 +12x is a concave down parabola. It's a parabola because the highest power of x is 2. and it is concave down because the co-efficient of x^2 is negative.
Now I am going to put it back into the previous form
y=x(12-x)
The roots are 0 and 12
The roots are where y=0, or to put it another way they are the x intercepts (where the curve crosses the x axis)
The vertex (if you want to find it and you really don't need to) will have an x=(0+12)/2 = 6 SO y=6(12-6)=36
That is, the vertex is (6,36)
Now you have 3 points, that is enough to define a parabola, so draw the parabola.
Graphically, to solution to
0 = x(12-x) is the intersection of y=0 (the xaxis) and y=x(12-x)
you can read the intersection points off your graph. x=0 and x=12.

Complete the square for the following quadratic function:

y = -6x2 + 36x - 12

Then, graph the function