SOLUTION: I REALLY NEED HELP WITH THESE QUESTIONS, ITS DUE TOMORROW AND I CANT SEEM TO GET THE ANSWERS TO THESE PROBELMS.. THANKS! 1) y=-3(x+5)^2-4 a)Find the cordinates of the vertex

Question 24734: I REALLY NEED HELP WITH THESE QUESTIONS, ITS DUE TOMORROW AND I CANT SEEM TO GET THE ANSWERS TO THESE PROBELMS.. THANKS!
1) y=-3(x+5)^2-4
a)Find the cordinates of the vertex
b)Find the equation of the axis of symmetry
c)Find the equation
-
2) Write in vertex form
y=2x^2+12x-3
-
3)For the quadrate relation 2x^2+x-15=0
a)Determine the number of roots
b)Solve
-
4)Describe and transform y=x^2 to obtain the graph of y=1/2(x+6)^2+12
-
5)Graph the following quadratic below
a)y=(x+2)^2-1
b)y=-2(x+3)^2
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
#1: Change to the form y-k = 4p(x-h)^2
which is a parabola opening up.
your equation becomes y+4=-3(x+5)^2
Vertex is h=-5; k=-4
Axis of symmetry is x=-5
#2: y=2(x^2+6x)-3
Create the form y-k=4p(x-h)^2
y+3= 2(x^2+6x+9)-18
y+21=2(x+3)^2
This is the "vertex form" you want.
#3: Use the determinant (b^2-4ac) to see how many real
number roots you have. Find the roots by using the
quadratic equation.
#4: x^2 becomes (x+6)^2 when you shift "6" to the left
Multiplying by 1/2 will stretch the (x+6)^2
Adding 12 will give the (1/2)(x+6)^2 a vertical shift up.
#5: Rewrite as y+1=(x+2)^2
which has vertex at (-2,-1)
I'll leave the last part of #5 to you.
Cheers,
Stan H.

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