what is the equation of the axis of symmetry of the graph of y=2x^2+12x-7?

Answer 1

You need to complete the square to figure the axis.

First factor out a 2 from 2x² and 12x:

y = 2(x² + 6x) - 7

Now complete the square by taking the coefficient on the x-term (6), dividing it in half (3) and squaring it (9). Now add and subtract this inside the parentheses:

y = 2(x² + 6x + 9 - 9) - 7

The first part of that is now a perfect square:
y = 2[ (x + 3)² - 9 ] - 7

Get the -9 out of the brackets by multiplying the 2 through:
y = 2(x + 3)² - 18 - 7
y = 2(x + 3)² - 25

Now you are in vertex form:
y = a(x - h)² + k

So the vertex is at the point (-3, -25). That makes the axis of symmetry where x = -3

Once you know the method of how it is derived, you can take a shortcut. The shortcut is to figure x = -b/2a. In your case a = 2, b = 12, so the axis of symmetry is x = -12/4 = -3

Answer 2

it's -3. just find the vertex and take the x value and set it equal to y and that will be your axis of symmetry (in this case it is y=-3)
or you can graph this equation in graphing calculator and find the minimum y value, (just in case you don't know that will be the vertex) so it goes back to what i just told you above that once you have the vertex you can find the axis of symmetry.

Answer 3

use the equation to find the vertex

x=-b/2a

then the axis of symetry is the answer to that equation

so:

x = -b/2a
=-12/4
=-3
therefore the equation of the axis of symetry of y=2x^2+12x-7 is x= -3

hope this helps :D