\begin{tabular}{|l|l|}
\hline Therefore the vertex is $(-1,-4)$ & Credit Recove \\
\hline
\end{tabular}
Question 3 Determine the vertex and x-intercepts of the parabola $f(x)=-2 x^{2}+4 x+5$
Answer
Final Answer: The vertex of the parabola is at \(\boxed{(1, 7)}\) and the x-intercepts are at $x = \boxed{1 - \frac{\sqrt{14}}{2}}$ and $x = \boxed{1 + \frac{\sqrt{14}}{2}}$. Therefore, the vertex and x-intercepts of the parabola $f(x)=-2 x^{2}+4 x+5$ are \(\boxed{(1, 7)}\) and \(\boxed{1 - \frac{\sqrt{14}}{2}, 1 + \frac{\sqrt{14}}{2}}\) respectively.
Step 1 ：Given the parabola $f(x)=-2 x^{2}+4 x+5$, we are asked to find the vertex and x-intercepts.
Step 2 ：The vertex of a parabola given in the form $f(x) = ax^2 + bx + c$ is given by the formula $(-\frac{b}{2a}, f(-\frac{b}{2a}))$.
Step 3 ：Substituting the values of $a$ and $b$ from the equation of the parabola, we find the x-coordinate of the vertex to be $1.0$.
Step 4 ：Substituting $x=1.0$ into the equation of the parabola, we find the y-coordinate of the vertex to be $7.00000000000000$.
Step 5 ：Therefore, the vertex of the parabola is at $(1, 7)$.
Step 6 ：The x-intercepts of the parabola are the solutions to the equation $f(x) = 0$.
Step 7 ：Solving this equation, we find the x-intercepts to be $1 - \frac{\sqrt{14}}{2}$ and $1 + \frac{\sqrt{14}}{2}$.
Step 8 ：Final Answer: The vertex of the parabola is at \(\boxed{(1, 7)}\) and the x-intercepts are at $x = \boxed{1 - \frac{\sqrt{14}}{2}}$ and $x = \boxed{1 + \frac{\sqrt{14}}{2}}$. Therefore, the vertex and x-intercepts of the parabola $f(x)=-2 x^{2}+4 x+5$ are \(\boxed{(1, 7)}\) and \(\boxed{1 - \frac{\sqrt{14}}{2}, 1 + \frac{\sqrt{14}}{2}}\) respectively.
