The vertex in this case is where the maximum or minimum (extreme points) are located.
The calculus definition of the vertex is when the derivative of the function reaches zero, when the instantaneous slope is not changing. The two answers are complimentary, because when the slope stops changing, it can do two things: either keep going or change directions. Because the function is a parabola (when the exponent is even), the function changes direction and thus a minimum or maximum (lowest or highest point compared to the points around it) is born.
In other words:
Think of it as the top of a mountain or bottom of a valley as you go straight over them.
This particular parabola has a negative coefficient, so it points downward, creating a maximum
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Interestingly enough this one has 3 ways of being solved. (There are almost always 3 ways, but the third way is usually not great without a really convenient function)
First: Use your graphing calculator to find the max.
Second: derive the equation (if you know the calculus, but this one feels like algebra)
-3x^2-4x-1
f ' = -6x-4=0
6x = -4
x= -2/3
Third: Factor / set the quadratic equation on the thing.
-3x^2-4x-1 = (-3x-1)*(x+1)
set each side to 0 to find the x-values that equal 0
-3x-1=0
x=-1/3
x+1=0
x=-1
Because our parabola (like all who have a degree (largest exponent in the function) of 2) is simple, the vertex is right between the two points where it crosses the x axis.
so (-1 + -1/3)/2 = -2/3
Note: Third method doesn't work with parabolas that don't cross the x axis, so I don't think that will ever be asked in the context of the class.
Also, quadratic equation (takes too long to type solution ((-b +- sqrt(b^2-4ac))/2a ) works the same way as the factoring, but should only be used when factoring isn't available (otherwise, its basically using a shotgun to kill a coach roach)
