a) Define a parable $$f(x)$$ with all its points in the first and second quadrant ($$\forall {x} \in \mathbb{R} , f(x)\geq 0$$) and a point $$(x_0,y_0)$$ placed in the third or fourth quadrant ($$y_0<0$$).
b) Find it the tangent straight lines to $$f(x)$$ which goes through that point. How many tangents can you find?
Development:
a) We could use $$f(x)=x^2+3$$, and the point $$(x_0,y_0)=(-2,-3)$$.
b) First, we use the expression of a general straight line going through the point $$(x_0,y_0):$$ $$$y-y_0=m\cdot(x-x_0)$$$ $$$y+3=m\cdot (x+2)$$$
Then we take the derivative of $$f(x)$$ and we obtain a generic tangency point parameterized by $$a$$: $$$f'(x)=2x$$$ $$$\mbox{tangency point}\equiv(a,a^2+3)$$$
The slope of the tangent straight line will be given by $$m=2\cdot a$$, and that straight line will go through the touching point $$(a,a^2+3)$$. Substituting in the equation for the tangent straight line: $$$a^2+3+3=2a\cdot (a+2) \Rightarrow a^2+4a-6=0 \Rightarrow a=-2\pm\sqrt{10}$$$
We should note that there are to possible values for the slope $$(m=2\cdot a):$$ $$$m_1=-4+2\sqrt{10}$$$ $$$m_2=-4-2\sqrt{10}$$$
This means that there will be two tangent straight lines to $$f(x)$$ and that they go through $$(x_0,y_0):$$
$$$\begin{eqnarray} & y_1=(-4+2\sqrt{10})\cdot(x+2)-3 & & y_1=(-4+2\sqrt{10})\cdot x-11+4\sqrt{10} \\\\ & & \Rightarrow & \\\\ & y_2=(-4-2\sqrt{10})\cdot(x+2)-3 & & y_2=(-4+2\sqrt{10})\cdot x-11-4\sqrt{10} \end{eqnarray}$$$
Solution:
a) $$f(x)=x^2+3$$, $$(x_0,y_0)=(-2,-3)$$.
b)
$$y_1=(-4+2\sqrt{10})\cdot x-11+4\sqrt{10}$$
$$y_2=(-4+2\sqrt{10})\cdot x-11-4\sqrt{10}$$