To stretch a graph horizontally across x=0, you replace every instance of "x" with "x/2" (so basically x has to be twice as large to get the same y value)

We don't want our stretch at x=0, we want it at x=4

To do that, we first move our graph...

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To stretch a graph horizontally across x=0, you replace every instance of "x" with "x/2" (so basically x has to be twice as large to get the same y value)

We don't want our stretch at x=0, we want it at x=4

To do that, we first move our graph 4 units to the left, then do the stretch, then move it back. To translate our graph, we replace every "x" with "x+4" (so basically, x has to be 4 less than it used to to get the same y value)

Our original graph:

`y=(x^3)-4(x^2)+x+6`, `x=4`

`y= (x+4)^3-4(x+4)^2+(x+4)+6` , `x=0`

Expanding out our new equation, we get:

`y=(x^3+3x^2(4)+3x(16)+64)-4(x^2+2x(4)+16)+(x+4)+6`

`y=x^3+12x^2+48x+64-4x^2-32x-64+x+4+6`

`y=x^3+8x^2+17x+10`

Now we stretch horizontally by a factor of 2, so we replace the "x"s in the equation with "x/2" :

`y=(x/2)^3+8(x/2)^2+17(x/2)+10`, `x=0`

(Note: the above graph has the same scale as the previous ones, so you can see the stretch without scaling distortion, but is centered differently)

The equation simplifies to:

`y=(1)/(8)x^3+2x^2+(17)/(2)x+10`

Now finally we need to move the graph back, 4 units to the right. To do that we replace every instance of "x" with "x-4":

`y=(1)/(8)(x-4)^3+2(x-4)^2+(17)/(2)(x-4)+10`, `x=4`

Our final equation simplifies to:

`y=(1)/(8)x^3+(1)/(2)x^2-(3)/(2)x`