How do you find the vertex of the parabola: #y=-5x^2+10x+3#?

1 Answer
jigglypuff314
Aug 10, 2015

The vertex is #(1,8)#

Explanation:

The x point of the vertex #(x, y)# is located on the parabola's Axis of Symmetry.
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The Axis of Symmetry of a Quadratic Equation
can be represented by #x=-b/{2a}#
when given the quadratic equation #y=ax^2+bx+c#
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In this case, given that #y=-5x^2+10x+3#
we can see that #a=-5# and #b=10#

plugging this into #x=-b/{2a}#
will get us: #x=-10/{2*(-5)}#
which simplifies to #x=1#
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Now that we know the x value of the vertex point, we can use it to find the y value of the point!
Plugging #x=1# back into #y=-5x^2+10x+3#
we will get: #y=-5+10+3#
which simplifies to: #y=8#
~
so we have #x=1# and #y=8#
for the vertex point of #(x,y)#
therefore the vertex is #(1,8)#

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