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$f(x)$ turns at $(-2,-1)$ , so $f'(-2)=0\implies k=8$ Now check that $$f(x)=2x^2+kx+k_2$$ $(k_2$ is some constant$)$ Since $(-2,-1)$ is a point on $f(x)$, so $-1=8-16+k_2\implies k_2=7$ Also check that $k_2$ is the answer to $(b)$.
To address the first part of your question, the thing is that wolfram alpha simply just made $e^C$ into a constant called $c_1$. So yes you were right in that $e^{\frac{x^2}{2} + c_1} = e^{\frac{x^2}{2}} * e^c$. But if you think about it, no matter what the value of $c$ is, $e^c$ will always come out to be some constant because $c$ is not variable. I know it ...