The problem is f(x)=x^2-12x-1
I have all the rest of the problems done except where it says the equation of the line of symmetry is ????
I know the vertex is (6,-37), the maximum is 6 and I think that part of the equation is x= -b/2a
Is this correct?
For a parabola such as yours, the axis of symmetry is very simple.
All you have to do is set x = 'the x value of the vertex'
so here
x = 6 is the axis of symmetry.
BTW, your parabola has a minimum rather than a maximum since it opens upwards.
That minimum is -37, and it occurs when x = 6
(the -b/(2a) is used to find the x of the vertex
Thank you Reiny. The problem said what is the equation of the line of the symmetry so I guess I over thought things, lol. Trying to prepare for my final that is due in a couple days and my brain is on overload bad.
Yes, you are on the right track. To find the equation of the line of symmetry for a quadratic function in the form f(x) = ax^2 + bx + c, you can use the formula x = -b/(2a).
In your case, the quadratic function is f(x) = x^2 - 12x - 1. To determine the equation of the line of symmetry, you need to identify the values of a and b from the quadratic function.
Comparing the function to the standard form ax^2 + bx + c, you can see that a = 1 and b = -12.
Now, plug these values into the formula x = -b/(2a):
x = -(-12)/(2*1)
x = 12/2
x = 6
Therefore, the equation of the line of symmetry is x = 6.