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

Aug 10, 2015

The vertex is $\left(1 , 8\right)$

Explanation:

The x point of the vertex $\left(x , y\right)$ 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 $x = - \frac{b}{2 a}$
when given the quadratic equation $y = a {x}^{2} + b x + c$
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In this case, given that $y = - 5 {x}^{2} + 10 x + 3$
we can see that $a = - 5$ and $b = 10$

plugging this into $x = - \frac{b}{2 a}$
will get us: $x = - \frac{10}{2 \cdot \left(- 5\right)}$
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 $y = - 5 {x}^{2} + 10 x + 3$
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 $\left(x , y\right)$
therefore the vertex is $\left(1 , 8\right)$