Solution-Slope intercept from, The line that is tangent to

Find an equation for the tangent line to the given curve at the point where x=x0

y=(x^2+3x-3)(2-2x) x0=1

The line that is tangent to y=(x^2+3x-3)(2-2x) at x0=1, in slope intercept from, is

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Slope intercept from : The line that is tangent to y=(x^2+3x-3)(2-2x) at x0=1, in slope intercept from, is

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