Question

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

Answer 1

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`
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so we have `x=1` and `y=8`
for the vertex point of `(x,y)`
therefore the vertex is `(1,8)`