How do you find the axis of symmetry and vertex point of the function: #Y=3x^2 + 8x + 4#?

1 Answer
Meave60
Oct 23, 2015

The axis of symmetry is #-4/3#.
The vertex is #(-4/3,-4/3)#.

Explanation:

#y=3x^2+8x+4# is a quadratic equation in the form #ax^2+bx+c#, where #a=3, b=8, c=4#.

Axis of Symmetry
An imaginary vertical line that divides the parabola into to equal halves.

Equation: #x=(-b)/(2a)#

#x=(-8)/(2*3)=-8/6=-4/3#

Axis of symmetry is #x=-4/3#.

Vertex
The maximum or minimum point of the parabola. Since #a# is a positive number, the parabola opens upward, so the vertex is the minimum point.

The #x# value of the point is the axis of symmetry, #x=-4/3#. To find the #y# value, substitute #-4/3# for #x# in the equation and solve for #y#.

#y=3x^2+8x+4=#

#y=3(-4/3)^2+8(-4/3)+4=#

#y=3(16/9)-32/3+4=#

#y=48/9-32/3+4=#

The LCD is #9#. Make each term have a denominator of #9#.

#y=48/9-32/3*3/3+4*9/9=#

#y=48/9-96/9+36/9#

Add the numerators.

#y=(48-96+36)/9=#

#y=-12/9#

Simplify.

#y=-4/3#

The vertex is #(-4/3,-4/3)#

graph{y=3x^2+8x+4 [-10, 10, -5, 5]}

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