What is the the vertex of $y = 2 x^{2} + 8 x + 5$ $y=2x^2 + 8x + 5$ ?

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What is the the vertex of $y = 2 x^{2} + 8 x + 5$ $y=2x^2 + 8x + 5$ ?

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Answer 1 · 2016-01-08T07:44:23

(-2,-3)

Explanation:

Well there are numerous ways to solve this but I'll tell you the shortest one (at least according to me).

When you see a parabola of form $y = a x^{2} + b x + c$ $y=ax^2+bx+c$ the slope of its vertex is 0. We know that the formula of the slope of any instantaneous line is $\frac{d y}{d x}$ $dy/dx$ so

$d \frac{2 x^{2} + 8 x + 5}{d x} = 0$ $d(2x^2+8x+5)/dx=0$
On solving this we get $x = - 2$ $x=-2$

Put this in our original equation of parabola and $y = - 3$ $y=-3$

This coordinates of the vertex are (-2,-3)

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What is the the vertex of $y = 2 x^{2} + 8 x + 5$ $y=2x^2 + 8x + 5$ ?

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