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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
`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:
`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:
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