-3x^2-6x-9 how do i find the stationary points
$\displaystyle f(x) = -3x^{2}-6x-9$
First derivative:
$\displaystyle f'(x) = -6x -6$
Second derivative:
$\displaystyle f''(x) = -6$
So far, from the second derivative test, you know that there is one event, and it is a global maximum (which would make sense, as this is a sad parabola )
Setting the first derivative to zero with give us the location of the event.
Setting the first derivative:
$\displaystyle f'(x) = -6x -6$
$\displaystyle 0 = -6x - 6$
$\displaystyle 6 = -6x$
$\displaystyle x = -1$
Now, we plug this into our original equation $\displaystyle f(x)$:
$\displaystyle f(-1) = -3x^{2}-6x-9$
$\displaystyle f(-1) = -3+6-9$
$\displaystyle f(-1) = -6$
So, the maximum occurs at $\displaystyle (-1, -6)$ where the point has a gradient of 0.
Since this is posted in PRE-CALCULUS, I'm going to assume that Calculus can not be used.
I'd note that the graphs of quadratic functions (parabolas) are symmetric and the turning point is on the axis of symmetry.
To find the axis of symmetry, you can average any two x-values that have the same y-value (such as the x-intercepts). This will give you the x co-ordinate of the turning point. Substitute it back into your function to find the y-value.