y=x2−4x
y′=(x2−4x)′=2x−4 The slope of the tangent to the curveat the point (3,−3) is
slope1=m1=y′(3)=2(3)−4=2 The slope of the normal to the curve at the point (3,−3) is
slope2−m2=−m11=−21 The equation of the normal to the curve at the point (3,−3) is
y−(−3)=−21(x−3) The equation of the normal to the curve at the point (3,−3) in slope-intercept form is
y=−21x−23
x -intercept: y=0=>−21x−23=0=>x=−3
A(−3,0)
y -intercept: x=0=>y=−21(0)−23=>y=−23
B(0,−23)
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