algebra 2 help....question?

Question

Identify the vertex and the axis of symmetry of the parabola. Identify points corresponding to P and Q.

http://i3.photobucket.com/albums/y81/slipknotme/i0000039.gif

a.
(1, –2), x = 1;
P'(0, –3), Q'(3, –6)

b.
(–2, 1), x = –2;
P'(0, –3), Q'(3, –6)

c.
(1, –2), x = 1;
P'(2, –3), Q'(–1, –6)

d.
(–2, 1), x = –2;
P'(2, –3), Q'(–1, –6)

Answer 1

This is just a matter of picking points off the graph
The vertex is the top point of the parabola (in this case)
which occurs 1 line to the right, and two lines down from the origin.
The vertex is at (1, -2)
In this case, the axis of symmetry is the vertical line through the vertex, which is the line x=1

I'll leave the other two up to you. Just count gridlines and get the co-ordinates.

Answer 2

a.
(1, –2), x = 1;
P'(0, –3), Q'(3, –6)
vertex . . (1, –2) . . . passing. .P'(0, –3)
equation of parabola
(y+2) = k (y-1)² . . substitute (0, -3) . . solving for k
(-3+2) = k(0-1)²
k = -1
y+2 = - (y-1)² . . . this is the equation of parabola
. . . . . it is symmetrical to x = 1

b.
(–2, 1), x = –2;
P'(0, –3), Q'(3, –6)
y = a x² + b x + c . . substitute (-2,1)
1 = a(-2)² + b(-2) + c
4a - 2b + c =1 . . . . . equation 1

y = a x² + b x + c. . substitute (0,-3)
-3 = 0 + 0 + c
c = -3

y = a x² + b x + c. . substitute (3,-6) . . & c = -3
-6 = 9a + 3b - 3
9a + 3b = -3
3a + b = -1 . . . . . . . equation 3
4a - 2b = 4 . . . . . from equation 1
6a +2b = -2 . . . . . . . equation 3 x 2 . . . . adding
10 a = 2
a = 0.2
b = -1 - 3a = -1-3(0.2) = -1.6
c = -3
equation is . . . y = 0.2 x² - 1.6x - 3
parabola with vertex (4, -6.2) . . facing up

c.
(1, –2), x = 1;
P'(2, –3), Q'(–1, –6)
equation is . . . y = -x² + 2x - 3
parabola with vertex (1, -2) . . facing down

d.
(–2, 1), x = –2;
P'(2, –3), Q'(–1, –6)
equation is . . . y = 2x² - x - 9
parabola with vertex (0.25, -9.125) . . facing up

Answer 3

What does P'(0,-3) stand for, as opposed to P(0,-3)?

Answer 4

y = - (x - 1)^2 - 2
You have a grid which is 1 unit x 1 unit. Simply count the units to each point vertically and horizontally.