y^2=8(x-3)
how would you get the coordinates.
vertex = 3,0
then what?
let x=1, or -1, and solve for y. This is a parabola, if you are graphing, you will need several points to plot.
vertex = (3,0)
using x = 2
(2,4) (2,-4)
> +3
translation = (5,4) (5,-4)
Is this correct??
so the points for the graph are
vertex (3,0)
(5,4) and (5, -4)
if I use x = 2
correct?
is that correct??
Yes, you are on the right track to find the coordinates to graph the parabola defined by the equation y^2 = 8(x-3).
To find additional points on the graph, you correctly begin by using x = 2. Plugging this value into the equation, you get:
y^2 = 8(2-3)
y^2 = 8(-1)
y^2 = -8
Since the square of a real number cannot be negative, this equation has no real solutions. Therefore, the point (2, -4) that you mentioned is incorrect.
However, you can still find other points by plugging in different x-values. For example, try using x = 4:
y^2 = 8(4-3)
y^2 = 8
Taking the square root on both sides, you get:
y = ±√8
y = ±2√2
So the points (4, 2√2) and (4, -2√2) are additional points on the graph.
In summary, the correct points to graph the parabola y^2 = 8(x-3) are:
- Vertex: (3, 0)
- Additional points: (5, 2√2), (5, -2√2), (4, 2√2), (4, -2√2)
Make sure to plot these points and connect them to form the parabolic shape.