TQSt Find the real zeros, if any, of the quadratic function using the quadratic formula. What are the \( x \)-intercepts, if any, of the graph of the function? ' \( f(x)=x^{2}+6 x+6 \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed) 1 poitis) posstile A. The zeros and the \( x \)-intercepts are the same. They are B. The zeros and the \( x \)-intercepts are different. The zeros are C. There is no real zero solution and no \( x \)-intercept.

To find the real zeros of the quadratic function \( f(x) = x^2 + 6x + 6 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 6 \), and \( c = 6 \). Plugging in these values gives us: \[ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(6)}}{2(1)} \] Calculating inside the square root: \[ x = \frac{-6 \pm \sqrt{36 - 24}}{2} = \frac{-6 \pm \sqrt{12}}{2} \] Simplifying further: \[ x = \frac{-6 \pm 2\sqrt{3}}{2} = -3 \pm \sqrt{3} \] Thus, the zeros and \( x \)-intercepts are: \(-3 + \sqrt{3}, -3 - \sqrt{3}\) So, the correct choice is: A. The zeros and the \( x \)-intercepts are the same. They are \(-3 + \sqrt{3}, -3 - \sqrt{3}\). The quadratic formula not only helps find zeros, but it's like a magic key for unlocking the mysteries of parabolas! You just need to plug in values of \( a \), \( b \), and \( c \) into the formula. Voila! Understanding the nature of the discriminant (the part under the square root) is crucial. If it's positive, you've got two distinct real zeros; if it's zero, you’ve got a single real zero or a "double root." If negative, well, no real zeros for you!