Question

need help please. I don't understand.

Draw the graph of relation defined by y^2=3x
-first, find the vertex of the equation
-select 1 pt above the axis of symmetry and 1 pt below
thanks !!

Answer 1

it's a parabola with vertex at the origin, opening towards the positive x axis...

Answer 2

i still don't understand. vertex is (0,0) . how do i get the other points ?

Answer 3

pick any x coordinate.... say x=1/3. then y = ???

Answer 4

for any positive x value you pick, you will have two y-values...
for example, when x=1/3, then
\(\large y^2=3(\frac{1}{3})\rightarrow y=\pm1 \)

here are your two points that correspond with x=1/3: (1/3, 1) and (1/3, -1)
keep doing that for several more x values and you'll see you have a parabola opening towards the right with the vertex at the origin.

Answer 5

got it ! thanks !