# find vortex, find zeros, and sketch graph, Quadratics,parabolas Y=2x^2+12x+10

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To find the vertex, you can use the formula "x = -b/2a" to find the x coordinate of the vertex, a, b, and c coming from the equation "y = ax^2 + bx + c". So, a = 2, b = 12, c = 10. So, for the x coordinate:

x = -12/(2*2) = -3

So, to find the y, we plug that back into the equation:

y = 2(-3)^2 + 12(-3) + 10

= 18 + (-36) + 10 = -8

So, the vertex = (-3,-8)

To find the zeros, we plug in 0 for y. So, we solve:

0 = 2x^2 + 12x + 10

Solving that equation for x:

0 = (2x + 2)(x + 5)

2x+2 = 0 and x+5 = 0

x = -1 and x = -5

Or, the zeros would be (-1,0) and (-5,0)

So, graphing the equation:

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steveschoen, did a wonderful job explaining how to get the answer without a calculator.

Now i'll show you another method of solving this problem

This method will be putting it as is in the calculator. (you should get this graph)

The places where the parabola goes through the x axis are the roots/zeros, and the vertex is the lowest or highest point:

therefore the zeros are x = -5 (-5,0) and x = -1 (-1,0)

and the vertex is (-3,-8)