Parabolas help!!! 9th grade maths?
I have this parabola function: y= (x+3) ^2
I need to find the zero points, which means I need too do this
0= (x+3)^2
How do I solve this?
Update:so the value to make the equation= 0 is -3 ?
there aren't two answers beacause the parabola is on the axis, so there's only one
Thank you so so much!!
Comments
0 = (x+3)^2
use FOIL method multiplying first, outer, inner, last terms together when you expand the squared.
But in this case you could just take the square root of both sides and get 0 = x + 3
0 = (x + 3)(x + 3)
0 = x^2 + 6x + 9
0 = (x + 3)(x + 3)
0 = x + 3
when y = 0, x is -3
(x + 3)² = 0
x + 3 = 0
x = -3
Well the only way that (x + 3)^2 can equal 0 is if x = - 3 so there is only one zero point in this case and that is when x = - 3.
(x + 3)^2 = 0
x + 3 = 0
x = -3
The vertex lies on the x axis: (-3, 0)
Indeed. What value do you have to give to "x" to make the equation [= 0] true?
Neat Jenis - the SQr of both sides!
But why are there not two answers?
This parabola is touching at (-3,0) not cutting x-axis.