Question

Please Check my Answer.

Calculate the discriminant and use it to determine how many real-number roots the
equation has.

3x^2 - 6x + 3 = 0

three real-number roots

two real-number roots <--- my answer

one real-number root

no real-number roots



9. Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis.

y = 2x2 + x - 3

1 point in common; vertex on x-axis

2 points in common; vertex below x-axis <---my answer

2 points in common; vertex above x-axis

no points in common; vertex below x-axis

Answer 1

Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis.

y = 3x^2 - 12x + 12


1 point in common; vertex on x-axis

2 points in common; vertex below x-axis

2 points in common; vertex above x-axis

no points in common; vertex above x-axis <----- My answer

Answer 2

I Just need these Three :S Please Check my answers :D

Answer 3

1. the discriminant is b^2 - 4ac, which here is 36-36 = 0. So there is one real-number root. (you were wrong)

2. logically: the a-coefficient is positive, so the parabola opens upwards. and it isn't a perfect square trinomial, so the first option is out. The third option is out, since it cannot have a minimum above the x-axis but 2 intercepts. The fourth one is out, because if the vertex is below the x-axis, then there must be 2 intersections. So it has to be the third one. (you were right)

3. We can't do the same logical argument here, so we have to find the vertex w/ brute math. the vertex's x-value is -b/2a, which here is 12/6 = 2. Now plug 2 in, and you get a vertex at (2 , 0), which is on the x-axis. So you get "1 point in common; vertex on x-axis" as your answer. (sorry, but your were wrong again)

Answer 4

Thank you SO Much :D