For the quadratic equation y=(x-3)^2-2. Determine: A) whether the graph opens up or down; B) the y-intercept; C)the x-intercepts; D) the points of the vertex. and D) graph the equation with at lease 10 pairs of points.

Answer 1

Vertex Form: Y = a(x-h)^2 + k.

A. The parabola opens upward because "a"
is positive.

B. Y = (x-3)^2 - 2.
Let X = o and solve for Y:
Y = (0-3)^2 - 2 = 7. = Y-int.

C. Let Y = 0 and solve forX:
Y = (x-3)^2-2 = 0.
(x-3)^2 = 2
Take sgrt of both sides:
x-3 = +-sqrt2
X = +-1.414 + 3 = 4.14, and 1.59. = X-intercepts.

D. V(h,k) = V(3,-2).

E. Select values of X below and above
h and calculate the corresponding value of Y.

(x,y). Y = (x-3)^2 - 2.
(-2,23)
(-1,14)
(0,7)
(1,2)
(2,-1)
V(3,-2)
(4,-1)
(5,2)
(6,7)
(7,14)
(8,23).

Answer 2

for the quadratic equation y=(x-3)^2-2 determine A) weather the graph opens up or down B) the y- intercept; C) the x intercept D) the points of the vertex and E ) graph the equation with at least 10 pairs of points