We proceed to differentiate: f'(x)=2x+4
Then we substitute for the point x0:
y−y0=f′(x0)∗(x−x0)y−y0=(2x0+4)∗(x−x0)y=y0+(2x)x0−2x02+4x−4x0y=x02+4x0−5+(2x)x0−2x02+4x−4x0y=2x(x0+2)−x02−5=mx+b
This leads us to the conclusion that the standard form for the tangent line is the line with slope m=2(x0+2) and the intersection at b=−(x02+5), where x0 is the point where the tangent line is placed.
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