SOLUTION: I need help solving for y=-3x^2-6x-5. I need to solve and find coordinates to beable to graph this. I know you substitute 0 for x and y however I get lost when I have to solve for

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I need help solving for y=-3x^2-6x-5. I need to solve and find coordinates to beable to graph this. I know you substitute 0 for x and y however I get lost when I have to solve for       Log On


   



Question 260499: I need help solving for y=-3x^2-6x-5. I need to solve and find coordinates to beable to graph this. I know you substitute 0 for x and y however I get lost when I have to solve for the squared part. Thanks
Answer by dabanfield(803) About Me  (Show Source):
You can put this solution on YOUR website!
The equation is for a parabola. The standard form of a parabola is:
y = ax^2 + bx + c
The role of 'a'
If a> 0, the parabola opens upwards
if a< 0, it opens downwards.
The axis of symmetry
The axis of symmetry is the line x = -b/2a
For this parabola we have a = -3, b = -6 and c = -5.
So the parabola opens downward and is symmetric around the vertical line
x = -b/2a = -(-6)/(2*(-3)) = 6/-6 = -1.
The maximum value for y will occur when x = -1.
When x = -1, y = -3*(-1)^2 -6*(-1) - 5 = -3 + 6 - 5 = -2
To get the shape of the parabola choose values of x on either side of -1, say x = 0 and x = -2 and compute the corresponding values for y.
Use the quadratic formula to solve:
-3x^2 - 6x - 5 = 0.
It will turn out there are no real solutions.