A curve \( C \) has equation \( y=f(x) \) wherc \[ f(x)=2(x+1)(x-3)^{2} \] (a) Sketch a graph of \( C \). Show on your graph the coordinates of the points where \( C \) cuts or meets the coordinate axes.

To sketch the graph of the curve \( C \) with the equation \( y=f(x) \) where \( f(x)=2(x+1)(x-3)^{2} \), we need to find the coordinates of the points where the curve cuts or meets the coordinate axes. First, let's find the x-intercepts by setting \( y=0 \) and solving for \( x \): \[ 0 = 2(x+1)(x-3)^{2} \] Solving this equation will give us the x-coordinates of the points where the curve intersects the x-axis. Next, we will find the y-intercept by setting \( x=0 \) and solving for \( y \): \[ y = 2(0+1)(0-3)^{2} \] This will give us the y-coordinate of the point where the curve intersects the y-axis. Once we have these coordinates, we can plot the points on the graph and sketch the curve \( C \) passing through these points. Let's calculate the x-intercepts and the y-intercept to determine the coordinates of the points where the curve intersects the coordinate axes. Solve the equation by following steps: - step0: Solve for \(x\): \(0=2\left(x+1\right)\left(x-3\right)^{2}\) - step1: Multiply the terms: \(0=2\left(x-3\right)^{2}\left(x+1\right)\) - step2: Swap the sides: \(2\left(x-3\right)^{2}\left(x+1\right)=0\) - step3: Elimination the left coefficient: \(\left(x-3\right)^{2}\left(x+1\right)=0\) - step4: Separate into possible cases: \(\begin{align}&\left(x-3\right)^{2}=0\\&x+1=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=3\\&x=-1\end{align}\) - step6: Rewrite: \(x_{1}=-1,x_{2}=3\) Solving the equation \( 0 = 2(x+1)(x-3)^{2} \) gives us the x-intercepts at \( x=-1 \) and \( x=3 \). Now, let's find the y-intercept by setting \( x=0 \) in the equation \( y=2(x+1)(x-3)^{2} \). Solve the equation by following steps: - step0: Solve for \(y\): \(y=2\left(x-3\right)^{2}\left(x+1\right)\) - step1: Rewrite the expression: \(y=2\left(x-3\right)^{2}x+2\left(x-3\right)^{2}\) - step2: Expand the expression: \(y=2x^{3}-10x^{2}+6x+18\) Solving the equation \( y=2(x+1)(x-3)^{2} \) gives us the y-intercept at \( y=2x^{3}-10x^{2}+6x+18 \). Now that we have the x-intercepts at \( x=-1 \) and \( x=3 \), and the y-intercept at \( y=2x^{3}-10x^{2}+6x+18 \), we can plot these points on the graph and sketch the curve \( C \) passing through these points. The x-intercepts are at \( (-1, 0) \) and \( (3, 0) \), and the y-intercept is at \( (0, 18) \). Plotting these points and sketching the curve \( C \) passing through them will give us the graph of the curve \( C \) with the equation \( y=f(x) \) where \( f(x)=2(x+1)(x-3)^{2} \).