Question 880836
Find the derivative of the function at that point.
The value of the derivative is the slope of the tangent line at that point.
Tangent and normal lines are perpendicular to each other. 
Determine the slope of the normal line from the tangent line slope. 
Use the point slope form to determine the equation of the normal line.
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{{{y=(2x-1)(3x+5)}}}
{{{y=6x^2+7x-5}}}
{{{dy/dx=12x+7}}}
At {{{x=1}}},
{{{m[T]=dy/dx=12(1)+7}}}
{{{m[T]=19}}}
Perpendicular lines have slopes that are negative reciprocals,
{{{m[T]*m[N]=-1}}}
{{{19*m[N]=-1}}}
{{{m[N]=-1/19}}}
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When {{{x=1}}}, {{{y=(2-1)(3+5)=8}}}
{{{y-8=-(1/19)(x-1)}}}
{{{y-8=-(1/19)x+1/19}}}
{{{y-8=-(1/19)x+1/19}}}
{{{y=-(1/19)x+1/19+152/19}}}
{{{y=-(1/19)x+153/19}}}
{{{19y=-x+153}}}
{{{x+19y-153=0}}}
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{{{drawing(300,300,-4,8,-2,10,grid(1),circle(1,8,0.2),graph(300,300,-4,8,-2,10,6x^2+7x-5,-(1/19)x+153/19))}}}
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