Przypomnijmy:

Styczną do wykresu funkcji f w punkcie (x₀, f(x₀)) nazywamy prostą opisaną równaniem:

$y-f\left(x_0\right)=f{}^{\prime }\left(x_0\right)\left(x-x_0\right)$  

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a)

$f\left(x\right)=x^2+4x-7$  

$P\left(1,-2\right)\text{zatem}{x}_0=1,f\left(x_0\right)=-2$     

$f{}^{\prime }\left(x\right)=2x+4$  

$f{}^{\prime }\left(1\right)=2\cdot 1+4=6$  

 

Zatem otrzymujemy:

$y-\left(-2\right)=6\left(x-1\right)$  

$y+2=6x-6\mid -2$  

$y=6x-8$  

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b)

$f\left(x\right)=-2x^2+x+3$  

$P\left(-2,-7\right)\text{zatem}{x}_0=-2,f\left(x_0\right)=-7$  

$f{}^{\prime }\left(x\right)=-4x+1$  

$f{}^{\prime }\left(-2\right)=-4\cdot \left(-2\right)+1=9$  

 

Zatem otrzymujemy:

$y-\left(-7\right)=9\left(x-\left(-2\right)\right)$  

$y+7=9\left(x+2\right)$  

$y+7=9x+18\mid -7$  

$y=9x+11$  

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c)

$f\left(x\right)=\frac{4x+2}{x+3}$  

$P\left(x_0,2\right)\text{zatem}f\left(x_0\right)=2$  

$2=\frac{4x_0+2}{x_0+3}$  

$2\left(x_0+3\right)=4x_0+2$  

$2x_0+6=4x_0+2\mid -2-2x_0$  

$4=2x_0\mid :2$  

$2=x_0$  

 

$f{}^{\prime }\left(x\right)=\frac{\left(4x+2\right){}^{\prime }\left(x+3\right)-\left(4x+2\right)\left(x+3\right){}^{\prime }}{{\left(x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{4\left(x+3\right)-\left(4x+2\right)}{{\left(x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{4x+12-4x-2}{{\left(x+3\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{10}{{\left(x+3\right)}^2}$  

$f{}^{\prime }\left(x_0\right)=\frac{10}{{\left(2+3\right)}^2}$  

$f{}^{\prime }\left(x_0\right)=\frac{10}{25}$  

$f{}^{\prime }\left(x_0\right)=\frac{2}{5}$  

 

Zatem otrzymujemy:

$y-2=\frac{2}{5}\left(x-2\right)$  

$y=\frac{2}{5}x-\frac{4}{5}+2$  

$y=\frac{2}{5}x+1\frac{1}{5}$  

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d)

$f\left(x\right)=\frac{3-x}{2x+18}$  

$P\left(x_0,\frac{1}{2}\right)\text{zatem}f\left(x_0\right)=\frac{1}{2}$  

$\frac{1}{2}=\frac{3-x_0}{2x_0+18}$  

$2x_0+18=2\left(3-x_0\right)$  

$2x_0+18=6-2x_0\mid +2x_0-18$  

$4x_0=-12\mid :4$  

$x_0=-3$  

 

$f{}^{\prime }\left(x\right)=\frac{\left(3-x\right){}^{\prime }\left(2x+18\right)-\left(3-x\right)\left(2x+18\right){}^{\prime }}{{\left(2x+18\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{-1\left(2x+18\right)-\left(3-x\right)\cdot 2}{{\left(2x+18\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{-2x-18-6+2x}{{\left(2x+18\right)}^2}$  

$f{}^{\prime }\left(x\right)=\frac{-24}{{\left(2x+18\right)}^2}$  

$f{}^{\prime }\left(-3\right)=\frac{-24}{{\left(2\cdot \left(-3\right)+18\right)}^2}$  

$f{}^{\prime }\left(-3\right)=\frac{-24}{{\left(-6+18\right)}^2}$  

$f{}^{\prime }\left(-3\right)=\frac{-24}{144}$  

$f{}^{\prime }\left(-3\right)=-\frac{1}{6}$  

 

Zatem otrzymujemy:

$y-\frac{1}{2}=-\frac{1}{6}\left(x-\left(-3\right)\right)$  

$y=-\frac{1}{6}\left(x+3\right)+\frac{1}{2}$  

$y=-\frac{1}{6}x-\frac{1}{2}+\frac{1}{2}$  

$y=-\frac{1}{6}x$