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

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

(1) Multiplying by -2 reflects the graph over the horizontal axis and performs a vertical stretch of factor 2.

(2) Subtracting 3 inside the function shifts the graph 3 units to the right.

(3) Adding 1 to the function shifts the...

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

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

(1) Multiplying by -2 reflects the graph over the horizontal axis and performs a vertical stretch of factor 2.

(2) Subtracting 3 inside the function shifts the graph 3 units to the right.

(3) Adding 1 to the function shifts the graph up 1 unit.

The graph shows the base function (black), the reflection (purple), the stretch (blue), the horizontal translation (green), and the vertical translation in red which is the graph you want.

Alternatively, recognize that this is in vertex form; `y=a(x-h)^2+k` where the vertex is at (h,k). In this case the vertex is (3,1). The parabola opens down, narrower than the base function. Starting at (3,1), another point can be located by going 1 unit left or right of the vertex and going down (since the parabola opens down due to the negative leading coefficient) 2 units; starting at that point another point can be found by moving left (or right if this is the right side of the graph) and going down 6 units. This gives you 5 points.

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

When x = 0 then f(x) = -17

When f(x) = 0;

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

`(x-3)^2 = 1/2`

`(x-3) = +-sqrt(1/2)`

`x = 3+-sqrt(1/2)`

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

` 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 f(x) = 1

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

- When x = 0 then y = 1
- y = 0 when `x = 3+sqrt(1/2)` and `x = 3-sqrt(1/2)`
- f(x) has a maximum at x = 3

Using this data we can plot the graph.

The graph is shown below.