# At what point on the graph f(x)=2xe^x is the tangent parallel to the line with equation y=2x-12?I have done this so far, no idea what to do next. y=2x-12 12=2x x=6

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You need to remember that a line is parallel to other line if they have like slopes.

You should remember that the slope of line is derivative of function at a point, hence the first derivative of function needs to be equal to 2 such that:

`f'(x) = 2`

You need to differentiate the function with respect to x using product rule such that:

`f'(x) = 2e^x + 2x*e^x`

You need to solve the equation `f'(x) = 2` such that:

`2e^x + 2x*e^x = 2`

Dividing by 2 both sides yields:

`e^x + x*e^x = 1 =gt e^x*(x+1) - 1 = 0`

You should notice that substituting 0 for x yields:

`e^0*(0+1) - 1 = 1*1 -1 =gt 0=0`

You should substitute 0 for x in equation of function to find the point you need such that:

`f(0) = 2*0*e^0 = 0`

**Hence, the tangent line to the graph of function is parallel to the line `y = 2x - 12` at origin (0,0).**