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 $$g(x)$$.
Development:
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 $$r(x)$$: $$$y+\dfrac{111}{16}=\dfrac{1}{2}\cdot\Big(x+\dfrac{7}{4}\Big)$$$ $$$r(x)=\dfrac{1}{2}\cdot x-\dfrac{97}{16}$$$
Solution:
a) $$f(x)=x^2+4x-3$$, $$g(x)=-2x+5$$.
b) $$r(x)=\dfrac{1}{2}\cdot x-\dfrac{97}{16}$$