# Compose the equation of the tangent and the equation of the normal to the graph of the function y=x^(2)-5x+4 if the abscissa of the tangent point x_(0)=-1

Compose the equation of the tangent and the equation of the normal to the graph of the function $y={x}^{2}-5x+4$ if the abscissa of the tangent point ${x}_{0}=-1$
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Francis Blanchard
$y={x}^{2}-5x+4$
${x}_{0}=-1,{y}_{0}=y\left({x}_{0}\right)=y\left(-1\right)=\left(-1{\right)}^{2}-5\ast \left(-1\right)+4=1+5+4=10$
${y}^{\prime }=\left({x}^{2}-5x+4{\right)}^{\prime }=2{x}^{2-1}-5+0=2{x}^{1}-5=2x-5$
${y}^{\prime }\left({x}_{0}\right)={y}^{\prime }\left(-1\right)=2\ast \left(-1\right)-5=-2-5=-7$
tangent equations:
$y-{y}_{0}={y}^{\prime }\left({x}_{0}\right)\ast \left(x-{x}_{0}\right)$
$y-10=-7\ast \left(x-\left(-1\right)\right)$
$y-10=-7\left(x+1\right)$
$y-10=-7x-7$
$7x+y-3=0$
normal equations:
$y-{y}_{0}=-\frac{1}{{y}^{\prime }\left({x}_{0}\right)}\ast \left(x-{x}_{0}\right)$
$y-10=-\frac{1}{-7}\ast \left(x-\left(-1\right)\right)$
$y-10=\frac{1}{7}\ast \left(x+1\right)$
$7y-70=x+1$
$x-7y+71=0$
Answer: tangent $7x+y-3=0$, normal $x-7y+71=0$