# Thread: Graph equation by hand

1. ## Graph equation by hand

y=-1/3(x-2)squared+1
I know the co-ordinates of the vertex are (2,1), I just do not know how to graph this by hand.

2. Hello,

You can do it this way :

- graph y=x². (I guess you know how to do it... ?)

- graph y=(x-2)² by translating the graph of x² of two units to the right (because x becomes x-2)

- graph y=-1/3*(x-2)² : multiplying by a constant. So the general appearance won't change, but because it's a negative constant, all the positive y will become negative, though the absolute value stays still. It's a symmetry to the x axis.

- graph y=-1/3*(x-2)²+1 : add 1 to the y, that is to say "elevate the curve of 1 unit"

Does it look clear to you ?

3. Find points where y=0.these two will lie on x axis.add them and divide it by 2 to get a third point.put this value in equation to get a particular point on y axis.now draw a rough sketch passing through these 3 points such that it make a U (distance between papallel lines will keep on increasing).I am giving a rough sketch so that you could check the nature your result..(the sketch that i am giving was not ment for your problem(you will get an inverted graph) but it will help you to understand the nature of graph).the point where graph take a turn is vertex.If still have any problem then ask.

4. I know that the sketching pattern for parabolas seems to be over 1 up 1 over 1 up 3 over 1 up 5, and so on, and if there is a number infront of the x(a), then do we not just multiply the sketch pattern from over 1 up one to over 1 up 1/3, and so on?

5. Hi!euclid2
the method that I have given is for general quadratic equation.
Other method for this particular question is graph shift.in this we just manipulate graph accordingly.
But the method that I gave previously is a general method for quadratic equations