Sketch the graph of \( f(x)=1+2 x-x^{2} \) and determine its domain and range.

I'm unable to plot the graph of the function \( f(x)=1+2x-x^{2} \) at the moment. However, I can help you determine the domain and range of the function. To find the domain of the function, we need to consider the values of \( x \) for which the function is defined. In this case, the function is a polynomial, and polynomials are defined for all real numbers. Therefore, the domain of the function is all real numbers, denoted as \( (-\infty, \infty) \). To find the range of the function, we need to consider the possible values that the function can take. Since the function is a quadratic function, it will have a maximum or minimum value. To find this value, we can complete the square or use calculus to find the vertex of the parabola. The range of the function will be all real numbers less than or equal to the maximum value or greater than or equal to the minimum value. Let's find the vertex of the parabola to determine the range of the function. Function by following steps: - step0: Find the local extrema: \(y=1+2x-x^{2}\) - step1: Use the appropriate notation: \(f\left(x\right)=1+2x-x^{2}\) - step2: Find the domain: \(f\left(x\right)=1+2x-x^{2},x \in \mathbb{R}\) - step3: Find the derivative: \(f^{\prime}\left(x\right)=2-2x\) - step4: Find the domain: \(f^{\prime}\left(x\right)=2-2x,x \in \mathbb{R}\) - step5: Substitute \(f^{\prime}\left(x\right)=0:\) \(0=2-2x\) - step6: Swap the sides: \(2-2x=0\) - step7: Move the constant to the right side: \(-2x=0-2\) - step8: Remove 0: \(-2x=-2\) - step9: Change the signs: \(2x=2\) - step10: Divide both sides: \(\frac{2x}{2}=\frac{2}{2}\) - step11: Divide the numbers: \(x=1\) - step12: Determine the intervals: \(\begin{align}&\left(-\infty,1\right),\left(1,+\infty\right)\end{align}\) - step13: Choose the points: \(x_{1}=0,x_{2}=2\) - step14: Find the values of the derivatives: \(f^{\prime}\left(0\right)=2,f^{\prime}\left(2\right)=-2\) - step15: The local maximum is at \(x=1:\) \(\begin{align}&f\left(x\right)=1+2x-x^{2},x=1\end{align}\) - step16: Calculate the function value: \(f\left(1\right)=2\) - step17: The local maximum of the function is \(2\) at \(x = 1:\) \(\begin{align}&\textrm{The local maximum is }2\textrm{ at }x = 1\end{align}\) The function \( f(x)=1+2x-x^{2} \) has a local maximum of 2 at \( x=1 \). Since the function is a quadratic function, it will have a minimum value at the vertex. The minimum value of the function will be the y-coordinate of the vertex. Therefore, the range of the function is all real numbers less than or equal to 2, denoted as \( (-\infty, 2] \).