A quadratic equation can be expressed either in standar or vertex form. It graph is a parabola.
The given equation in the problem is in vertex form:
`y = a(x-h)^2+k`
where (h,k) is the vertex.
The `a` determines the direction of the parabola. It opens up if it is positive and down if negative.
In the equation,
`y = -2(x+3)^2 - 1`
we can directly tell that its vertex is (-3,-1) and the parabola opens downward.
Since it is downward, the vertex is the maximum point.
The next step to graph the equation is to assign values of x that is to the left and right of the vertex. And, solve for y.
At the left of the vertex, let:
`x = -5` ,
`y = -2(-5+3)^2-1 = -2(-2)^2-1=-2*4-1=-8-1=-9`
And at the right of the vertex, let:
`y =-2(-1+3)^2 - 1 = -2(2)^2 - 1 = -2*4-1=-8 -1= -9`
Then, plot the vertex (-3,-1) and the other two points (-5,-9) and (5,-9). Aftewards, connect them. Note that when connecting the three points it should follow the shape of the parabola.
Hence the graph of `f(x) = -2(x+3)^2-1` is: