Question 1179671
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Note, for future reference: In typed text, use "^" to denote an exponent.  In your problem, "y=-2(x-3)^2+9" instead of "y=-2(x-3)2+9".<br>
(1) Vertex and axis of symmetry:  The given equation is in vertex form: y=a(x-h)^2+k<br>
In that form, the vertex is at (h,k).  And the axis of symmetry is the vertical line through the vertex.<br>
ANSWER: The vertex is (3,9); the axis of symmetry is x=3.<br>
(2) Maximum and minimum:  The coefficient on the x^2 term is negative, so the graph is a parabola that opens downward.  So it has a maximum but no minimum.  The maximum is at the vertex.
It's easy to calculate the maximum value when the equation is in vertex form -- it's at the vertex (i.e., when the (x-3)^2 part of the equation is equal to 0).
ANSWER: maximum value at (3,9); no minimum value.<br>
(3) y-intercept: When x=0. -2(-3)^2+9 = -2(9)+9 = -18+9 = -9.
ANSWER: y-intercept (0,-9)<br>
(4) zeros (x-intercepts): When y=0.  Put the equation in standard form.  If it can be factored, then finding the zeros is simple; if not, use the quadratic formula.
-2(x-3)^2+9 = -2(x^2-6x+9)+9 = -2x^2+12x-9
That does not factor, so use the quadratic formula.<br>