Question 602355: Hello,
would you pls check if my answer is correct?
1)Without drawing the graph of the equations, 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.
a. y = 3x^2 – 12x + 12
b. y = –2x^2 + x + 3
My answer : a. 1 point in common; vertex on x-axis
b. 2 points in common; vertex above x-axis
Thanks a lot,
H.H
P.S.:This is my 2nd submission for this question.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 1)Without drawing the graph of the equations, 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.
a. y = 3x^2 – 12x + 12
Find the discriminant
D = b^2 - 4ac
D = (-12)^2 - 4*3*12
D = 0 --> 1 real number solution
It's actually 2 solutions that are equal, but that makes one point on the x-axis.
-----------------
b. y = –2x^2 + x + 3
D = 1 - 4*-2*3
D = 25
D > 0 --> 2 real solutions, 2 crossings of the x-axis
------------
My answer : a. 1 point in common; vertex on x-axis
Correct.
============================================
b. 2 points in common; vertex above x-axis
2 points. The coefficient of the x^2 terms is negative, so it opens downward.
--> vertex above the x-axis
Correct.
|
|
|
