Question

Identify the vertex, axis of symmetry, maximum and minimum value, and range of y=3x^2 -6x+2

Can someone please help me find these?
Thanks in advance.

Answer 1

The coefficient of x² is 3 which is positive so that means it will have a curve that is concave up (shaped like a 'U'). It will have a minimum value and its maximum value will be infinity.

Answer 2

Okay, thanks. But how did you get that?

Answer 3

Just by knowing that the coefficient of x² is a positive number '3'.

Answer 4

I got...
vertex: x=- 3/2(-6) = 1/4
y=3(1/4)^2 -6(1/4)+2 = 11/16

axis of symmetry: x= 1/4
minimum value: y= 11/16 at x= 1/4
range: y (greater than or equal to) 11/16

Where did I go wrong?

Answer 5

It seems you'll also need the vertex which will also give you a minimum value. This is from calculus but gives you the correct answer.
We take the derivative of y=3x^2 -6x+2 and make it equal zero. The derivative is
6x -6 and 6x - 6 = 0 when x = 1. The minimum value is at x=1 and when x = 1, y=
3x² -6x +2 = 3-6+2 = -1
So minimum is at (1, -1)
The minimum occurs at (1, -1) so the axis of symmetry is a vertical line going through the point x = 1.

Answer 6

y = (3x)squared - 4x - 2

well you can do the completing the squares
so factor this to:
y = ax(sqared) + bx + c ==> y = a(x-h)^2 + k

y = (3x)squared - 4x +2
y = (3x^2 - 4x +4)+ 2 - 4 =============> ((1/2)(-4))) = (-2)^2 = +4
factor the inside!!!
Use quadratic do the rest srry dude to lazy D:

hope this helps

Answer 7

@Gracy_KSL @wolf1728 thanks both of you. So what exactly did I do wrong?

Answer 8

I'd guess that you used the vertex formula incorrectly. (I don't know it - I can look it up).

Answer 9

That's okay. Thanks so much!! :)

Answer 10

Okay the vertex formula for the x value is
x = -b/2a
In the quadratic equation 3x^2 -6x+2
a=3 b=-6 anc c=2
x = -(-6) / 2*3
x = 6/6
x=1

Answer 11

I'd guess you get the
y
value by plugging in the value of x =1 into the equation
= 3*1 -6*1 +2 = -1

Answer 12

Thank you so much. You were very helpful!!

Answer 13

Thanks Karebear
:-)