Help with graphing a Quadratic Equation: y = x(x - 9)

[D!ddy]

I need to graph the equation y = x(x - 9)

so far i know that the zeros are 0, and 9, but I don't know how to find the vertex, i know that I should use the complete the square formula, (when i turn it into y= x^2 - 9x)..but I'm a little lost since there's no constant term.. so if you could tell me how to find the vertex using the completing the square formula or another way I would really appreciate it

Thanks,
Mo

 

[Loren]

What happens when x is half way between 0 and 9?

 

[o_O]

Imagine the constant term is 0. So:
y = x(x - 9)
y = x² - 9x + 0

and complete the square normally.

 

[jwpaine]

x coordinate of virtex: -b/(2a)

 

[D!ddy]

to loren: i know that its a positive graph since the leading coefficient is positive but the vertex is below the x-axis.. also i tried to imagine to add 0 already but whats confusing me is where to put the negative signs..because if i keep it positive( y+(81/4)=x^2 - 9x + (81/4) i end up back where i was..i know I'm missing something..so please tell me where i went wrong..

 

[D!ddy]

sorry for double posting but looks like i jumped the gun..i skipped a step but finally found y=(x- 9/2)^2 - 81/4 and thats the vertex i found in my calculator...thanks for you hints though really helpful.

Thanks,
Mo

 

[o_O]

Completing the square:
y = x² + 9x
y = (x² + 9x + (9/2)²) - (9/2)²
etc. etc.