Question 412883
Graph the quadratic equation, and find the following
y=-(x-3)^2+4
Vertex:
X-intercepts (zeros):
Axis of symmetry equation:
y-intercept:

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Standard form of a parabola:y=(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. Parabola opens upward if coefficient of (x-h)^2 term is positive and downward if the coefficient is negatve.
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Given parabola opens downward.(by inspection)
Vertex: (3,4) (by inspection)
Axis of symmetry: x=3 (by inspection)
solving for y-intercept
let x=0
y=-9+4=-5
y-intercept=5
solving for x-intercepts
let y=0
(x-3)^2=4
x-3=+-sqrt(4)
x=3+-2
x=1
x=5

see the graph below:

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{{{ graph( 300, 300, -10, 10, -10, 10, -(x-3)^2+4) }}}