What is the axis of symmetry and vertex for the graph #y=3x^2+12x-2#?

2 Answers
Feb 14, 2018

Axis of symmetry: #x = -2#
Vertex: #(-2, -14)#

Explanation:

This equation #y = 3x^2 + 12x - 2# is in standard form, or #ax^2 + bx + c#.

To find the axis of symmetry, we do #x = -b/(2a)#.

We know that #a = 3# and #b = 12#, so we plug them into the equation.
#x = -12/(2(3))#

#x = -12/6#

#x = -2#

So the axis of symmetry is #x = -2#.


Now we want to find the vertex. The #x#-coordinate of the vertex is the same as the axis of symmetry. So the #x#-coordinate of the vertex is #-2#.
To find the #y#-coordinate of the vertex, we just plug in the #x# value into the original equation:
#y = 3(-2)^2 + 12(-2) - 2#

#y = 3(4) - 24 - 2#

#y = 12 - 26#

#y = -14#

So the vertex is #(-2, -14)#.

To visualize this, here is a graph of this equation:

enter image source here

Hope this helps!

Feb 14, 2018

Axis of Symmetry is the line #color(blue)(x=-2#

Vertex is at: #color(blue)((-2, -14).#It is a minimum.

Explanation:

Given:

#color(red)(y=f(x) = 3x^2+12x-2#

We use the Quadratic formula to find the Solutions:

#color(blue)(x_1, x_2=(-b+-sqrt(b^2-4ac))/(2a)#

Let us look at #color(red)(f(x)#

We observe that #color(blue)(a=3; b=12; and c=(-2)#

Substitute these values in our Quadratic formula:

We know that our discriminant #b^2-4ac# is greater than zero.

#color(blue)(x_1, x_2=[-12+-sqrt[12^2-4(3)(-2)])/(2(3))#

Hence, we have two real roots.

#x_1, x_2=[-12+-sqrt([144+24]])/(6)#

#x_1, x_2=[-12+-sqrt(168])/(6)#

#x_1, x_2=[-12+-sqrt(4*42])/(6)#

#x_1, x_2=[-12+-sqrt(4)*sqrt(42])/(6)#

#x_1, x_2=[-12+-2*sqrt(42])/(6)#

#x_1, x_2=[-12]/6+-(2*sqrt(42])/(6)#

#x_1, x_2=-2+-(cancel 2*sqrt(42])/(cancel 6 color(red)3)#

#x_1, x_2=-2+sqrt(42)/3, -2-sqrt(42)/3#

Using a calculator, we can simplify and get the values:

#color(blue)(x_1=0.160247, x_2=-4.16025#

Hence, our x-intercepts are: #color(green)((0.16,0),(-4.16,0)#

To find the Vertex,

we can use the formula: #color(blue)((-b))/color(blue)((2a)#

Vertex: #-12/(2(3)#

#rArr -12/6=-2#

This is our x-coordinate value of our Vertex.

To find the y-coordinate value of our Vertex:

Substitute the value of #color(blue)(x=-2# in

#color(red)(y = 3x^2+12x-2#

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

#y = 3(4)-24-2#

#y = 12-24-2 = 14#

Vertex is at: #color(blue)((-2, -14)#

The coefficient of the #color(green)(x^2# term is Positive and hence, our Parabola Opens Upward, and it has a minimum. Please refer to the image of the graph below to verify our solutions:

enter image source here

The Axis of symmetry of a parabola is a vertical line that divides the parabola into two congruent halves.

The Axis of Symmetry always passes through the Vertex of the Parabola. The #x# coordinate of the vertex is the equation of the Axis of Symmetry of the Parabola.

Axis of Symmetry is the line #color(blue)(x=-2#