\begin{tabular}{|l|l|l|l} \\ What is the domain of \( y=-3 x^{2}+6 x+4 ? \) \\ \hline\( y \leq 7 \) & \( x \leq 7 \) & \( x \leq 4 \) \\ \hline\end{tabular}

To find the domain of the function \( y = -3x^2 + 6x + 4 \), we need to determine the values of \( x \) for which the function is defined. Given the inequality \( y \leq 7 \), we can substitute the function \( y = -3x^2 + 6x + 4 \) into the inequality to find the domain of the function. Substitute \( y = -3x^2 + 6x + 4 \) into the inequality \( y \leq 7 \): \(-3x^2 + 6x + 4 \leq 7\) Simplify the inequality: \(-3x^2 + 6x - 3 \leq 0\) Now, we need to solve this quadratic inequality to find the domain of the function. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(-3x^{2}+6x-3\leq 0\) - step1: Rewrite the expression: \(-3x^{2}+6x-3=0\) - step2: Factor the expression: \(-3\left(x-1\right)^{2}=0\) - step3: Divide the terms: \(\left(x-1\right)^{2}=0\) - step4: Simplify the expression: \(x-1=0\) - step5: Move the constant to the right side: \(x=0+1\) - step6: Remove 0: \(x=1\) - step7: Determine the test intervals: \(\begin{align}&x<1\\&x>1\end{align}\) - step8: Choose a value: \(\begin{align}&x_{1}=0\\&x_{2}=2\end{align}\) - step9: Test the chosen value: \(\begin{align}&x<1\textrm{ }\textrm{is the solution}\\&x>1\textrm{ }\textrm{is the solution}\end{align}\) - step10: Include the critical value: \(\begin{align}&x\leq 1\textrm{ }\textrm{is the solution}\\&x\geq 1\textrm{ }\textrm{is the solution}\end{align}\) - step11: The final solution is \(x \in \mathbb{R}:\) \(x \in \mathbb{R}\) The solution to the quadratic inequality \(-3x^2 + 6x - 3 \leq 0\) is \( x \in \mathbb{R} \), which means that the domain of the function \( y = -3x^2 + 6x + 4 \) is all real numbers. Therefore, the domain of the function is \( x \in \mathbb{R} \).