Solution. Take out the coefficient of \(x^{2}\) from the two \(x\) terms, being careful with the sign: \begin {align*} f(x)&=-2x^{2}+x+5 =-2\left (x^{2}-\tfrac {1}{2}x\right )+5\\ &=-2\left [\left (x-\tfrac {1}{4}\right )^{2}-\tfrac {1}{16}\right ]+5 =-2\left (x-\tfrac {1}{4}\right )^{2}+\tfrac {1}{8}+5\\ &=-2\left (x-\tfrac {1}{4}\right )^{2}+\tfrac {41}{8}. \end {align*}

(a). The square is never negative and \(-2\) is negative, so the largest value of \(f\) occurs when the bracket vanishes, at \(x=\frac {1}{4}\). The turning point is \(\left (\frac {1}{4},\frac {41}{8}\right )\), and it is a maximum because \(a=-2<0\).

(b). The \(y\)-intercept is \(f(0)=5\). For the \(x\)-intercepts solve \(f(x)=0\): \[-2\left (x-\tfrac {1}{4}\right )^{2}=-\tfrac {41}{8} \implies \left (x-\tfrac {1}{4}\right )^{2}=\tfrac {41}{16} \implies x=\frac {1\pm \sqrt {41}}{4},\] approximately \(-1.35\) and \(1.85\). The curve is a downward parabola with vertex \(\left (\frac {1}{4},\frac {41}{8}\right )\), crossing the \(y\)-axis at \(5\) and the \(x\)-axis at those two points.

Solution. Complex roots means a negative discriminant. With \(a=1\), \(b=2-k\) and \(c=1-2k\), \[D=(2-k)^{2}-4(1-2k)=4-4k+k^{2}-4+8k=k^{2}+4k .\] The condition \(D<0\) gives \[k^{2}+4k<0 \implies k(k+4)<0 ,\] a product of two factors that is negative exactly when they have opposite signs, which happens for \[-4<k<0 .\]