How do you find the x-coordinate of the vertex for the graph $4x^2+16x+12=0$ ?

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How do you find the x-coordinate of the vertex for the graph $4x^2+16x+12=0$ ?

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Answer 1 · 2018-05-29T15:59:29

$x=-2$

Explanation:

Your quadratic equations is in standard form $ax^2+bx+c$ so the x-coordinate of the vertex (also called the axis of symmetry) is found with the formula:

$aos=(-b)/(2a)$

$4x^2+16x+12$

$aos=(-16)/(2*4) = -2$

graph{4x^2+16x+12 [-10, 10, -5, 5]}

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How do you find the x-coordinate of the vertex for the graph $4x^2+16x+12=0$ ?

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How do you find the x-coordinate of the vertex for the graph $4x^2+16x+12=0$ ?