(1)) Find all values of \( k \) that make \( f(x) \) continuous at \( x=1 \) \[ f(x)=\left\{\begin{array}{ll}-x^{2}-3 & \text { if } x \leq 1 \\ -5 k x+k^{2} & \text { if } x>1\end{array}\right. \] \( k= \) Write your answer as a single value or a list of values separated by commas.

To find the values of \( k \) that make the function \( f(x) \) continuous at \( x=1 \), we need to ensure that the left-hand limit and the right-hand limit of \( f(x) \) as \( x \) approaches 1 are equal. Given: \[ f(x)=\left\{\begin{array}{ll}-x^{2}-3 & \text { if } x \leq 1 \\ -5 k x+k^{2} & \text { if } x>1\end{array}\right. \] To find the left-hand limit: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (-x^{2}-3) = -1^{2}-3 = -4 \] To find the right-hand limit: \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (-5kx+k^{2}) = -5k(1)+k^{2} = -5k+k^{2} \] For the function to be continuous at \( x=1 \), the left-hand limit and the right-hand limit must be equal. Therefore, we set the two limits equal to each other: \[ -4 = -5k+k^{2} \] Now, we can solve this equation to find the values of \( k \) that make the function continuous at \( x=1 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(-4=-5k+k^{2}\) - step1: Swap the sides: \(-5k+k^{2}=-4\) - step2: Move the expression to the left side: \(-5k+k^{2}+4=0\) - step3: Factor the expression: \(\left(k-4\right)\left(k-1\right)=0\) - step4: Separate into possible cases: \(\begin{align}&k-4=0\\&k-1=0\end{align}\) - step5: Solve the equation: \(\begin{align}&k=4\\&k=1\end{align}\) - step6: Rewrite: \(k_{1}=1,k_{2}=4\) The values of \( k \) that make the function \( f(x) \) continuous at \( x=1 \) are \( k=1 \) and \( k=4 \).