Intermediate Algebra

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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.

  • Intermediate Algebra -

    y=(x-3)^2-2
    or
    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 )

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