# Find the vertex of parabola f(x)=6x^2+12x+78?

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### 2 Answers

The coordinates of the vertex of the parabola are:

V(-b/2a ; -delta/4a)

a,b,c are the coefficients of the quadratic:

y = ax^2 + bx + c

Comparing, we'll get the values of the coefficients:

a = 6, b = 12 , c = 78

We'll determine the x coordinate of the vertex:

xV = -b/2a

xV = -12/12

xV = -1

We'll determine the y coordinate of the vertex:

yV = -delta/4a

delta = b^2 - 4ac

delta = 144 - 4*6*78

delta = 144 - 1872

yV = -(-1728)/24

yV = 72

**The coordinates of the vertex of the parabola are: V(-1, 72).**

`f(x)=6x^2+12x+78`

`a=6` `b=12` `c=78`

use the formula `-b/(2a) ` to find the x value of the vertex

`x=-12/(2(6))` `x=-12/12` `x=-1`

plug in the x into the original problem

`y=6(-1)^2+12(-1)+78`

`y=6(1)-12+78`

`y= 6-12+78`

`6-12=-6+78=72`

`y=72`

the vertex is `(-1,72)`