What is the maximum value of y in (y= -3x^2-6x-1)

a. -2
b. -1
c. 1
d. 2

Please answer the question and explain how you got the answer :) <3 Oh and also, what does "maximum value" mean? Thanks <3 :)

Let's differentiate like a boss
Y'=-6x-6
Where is this graph zero?
6=-6x
X=-1

The maximum is at x=-1

-------->>>>>So y=2

The maximum is where the graph is the highest. If you graph it it'll be a parabola going down so it has one point its the highest..like a hill.

Betcha didn't think id use calculus. d:D

If you do not know Calculus, then use the following

find x so that y = -1 = -3x²-6x-1 (-1 is the constant of the quadratic polynomial)

The equation becomes 0 = -3x² - 6x = -3x(x + 2)

The solutions are x = 0 and also x = -2. The x-value of the vertex (maximum value of the function) is the average of these two numbers: (0 + -2)/2 = -1

The corresponding y is -3(-1)² - 6(-1) - 1 = -3 + 6 – 1 = 2

The maximum point is (-1, 2)

It works this way every time. Who needs Calculus???

ProfRay

The maximum value occurs at the vertex, You can use x = -b/(2a) (derived from calculus optimization) to find the x-coordinate of the vertex of a parabola.

x = -(-6)/(2*(-3))
x = 6/(-6)
x = -1

In order to find the value, plug this number in for x and solve for y.

y = -3(-1)² - 6(-1) - 1
y = -3 + 6 - 1
y = 2

y= -3x^2-6x-1
y'=-6x-6 to get the maximum value we make y'=0
6x=-6
x=-6/6
x=-1 now we get the maximum value
y=-3(-1)²-6(-1) -1
=-3+6-1
=2 so the answer is d. 2

y'=-6x-6
0=-6x-6
x=-1 so max occurs where x=-1
-3(-1)^2-6(-1)-1
-3+6-1
2 is max value of y

im sorry.. i have no idea, this is depressing!