Normal straight line to a curve at a point
a) Define two functions $f(x)$ and $g(x)$, the first one a parable (equation of the second degree) and the second one a straight line.
b) Find the straight line $r(x)$, tangent to $f(x)$ and normal to $g(x)$.
a) Two possible candidates are: $f(x)=x^2+4x-3$ and $g(x)=-2x+5$.
b) First we look for the slope of $r(x)$.
Slope of $g(x): \ g'(x)=-2$
$r(x)$ normal to $g(x) \rightarrow r'(x)=-\dfrac{1}{-2}=\dfrac{1}{2}$
Now we should look for the for the point where the derivative of $f(x)$ is $\dfrac{1}{2}$. This is $$f'(a)=2a+4=\dfrac{1}{2} \Rightarrow a=-\dfrac{7}{4}$$ $$f\Big(-\dfrac{7}{4}\Big)=\Big(-\dfrac{7}{4}\Big)^2+4\Big(-\dfrac{7}{4}\Big)-3=-\dfrac{111}{16}$$
The touching point will be $(a,f(a))=\Big(-\dfrac{7}{4},-\dfrac{111}{16} \Big)$
The equation of the straight line is written $r(x)$: $$y+\dfrac{111}{16}=\dfrac{1}{2}\cdot\Big(x+\dfrac{7}{4}\Big)$$ $$r(x)=\dfrac{1}{2}\cdot x-\dfrac{97}{16}$$
a) $f(x)=x^2+4x-3$, $g(x)=-2x+5$.
b) $r(x)=\dfrac{1}{2}\cdot x-\dfrac{97}{16}$