posted by Anonymous .
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.
y=+1(x-3)^2-2 vs y = a(x-p)^2 + q
since a=1 , a> 0 , so it opens upwards
y=intercept, let x = 0
y = (-3)^2 - 2 = 7
x-intercept, let y = 0
(x-3)^2 - 2 = 0
(x-3)^2 = 2
x-3 = ± √2
x = 3 ± √2
"points of the vertex " ????
the vertex IS a point, it does not have points, it has coordinates
they are (3,-2) , (from (p,q) )
For D) just pick any suitable values for x, evaluate the corresponding y's and plot.
(pick values of x immediately above and below +3 )