Answer: First Option
a) exponential function going through point (0, 2) and ending up on the right
Step-by-step explanation:
Look at the attached image, the red line represents a function of the form:

Note that this function cuts to the axis and at the point (0, 1)
Also when x tends to ∞ f(x) tends to ∞ and when f(x) tends to -∞ then f(x) tends to zero.
If we perform the transformation
then the graph of y is equal to the graph of f(x) displaced 1 unit up. Then the new cutting point with the axis y will be: (0, 2) as shown in the attached image (blue line)
The transform function is 
Finally the answer is the first option
I believe the answer the answer is 5.1 b
It is prime, it has no factors besides 1,23
\left[x _{2}\right] = \left[ \frac{-1+i \,\sqrt{3}+2\,by+\left( -2\,i \right) \,\sqrt{3}\,by}{2^{\frac{2}{3}}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{24}+\left( \frac{-1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{\sqrt[3]{2}}\right][x2]=⎣⎢⎢⎢⎢⎡2323√(432by+√(−6912+41472by+103680by2+55296by3))−1+i√3+2by+(−2i)√3by+3√224−3√(432by+√(−6912+41472by+103680by2+55296by3))+(24−1i)√33√(432by+√(−6912+41472by+103680by2+55296by3))⎦⎥⎥⎥⎥⎤
totally answer.