Graph `y=-2(x+3)^2-1` :

The base function is `y=x^2` .

(1) Multiplying by -2 reflects the base function across the x-axis, and stretches by a factor of 2.

(2) Adding 3 inside the function shifts the base function 3 units to the left.

(3) Subtracting 1 from the function shifts the graph 1 unit down.

To graph, take the base function (black) , reflect over the x-axis(purple), stretch by a factor of 2(blue), slide 3 units left(green), then 1 unit down(red) which is the graph you want

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`f(x) = -2(x+3)^2-1`

When x = 0 then f(x) = -19

When f(x) = 0;

`-2(x+3)^2-1 = 0`

          `(x+3) = +-sqrt(1/2)`

                ` x = -3+-sqrt(1/2)`

When f(x) = 0 then `x = -3+sqrt(1/2)` and `x = -3-sqrt(1/2)` .

`f(x) = -2(x+3)^2-1`

`f'(x) = -4(x+3)`

`f''(x) = -4`

When f'(x) = 0;

`-4(x+3) = 0`

          `x = -3`

At x=-3 then y = -1

Since f''(x)<0 we have a maximum at x = -3 for f(x)

• At x = 0 then f(x) = -19
• At f(x) = 0 then `x = -3+sqrt(1/2)` and `x = -3-sqrt(1/2)`
• f(x) have a maximum at x = -3
• Maximum of f(x) = -1

Using the data above we can plot the graph.

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:

`x=-1` ,  

`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: