\begin{tabular}{l}\( \qquad g(x)=x^{2}+8 x-6 \) \\ \( \qquad(-5)^{2}+8(-5)-6=25-40 \sim 6=-2 \) \\ Find the vallowing function in \( x \in \mathbb{R} \) : \\ \( \qquad g(-5) \) (c) \\ Use calculus to find the value of \( x \) which gives the minimum value of \( g(x) \) \\ \hline\end{tabular}

To find the value of the function \( g(x) \) at \( x = -5 \), we can double-check our calculations. Substituting \(-5\) into the function gives us: \[ g(-5) = (-5)^{2} + 8(-5) - 6 = 25 - 40 - 6 = -21. \] So, \( g(-5) = -21 \). To find the minimum value of \( g(x) = x^{2} + 8x - 6 \) using calculus, we take the derivative \( g'(x) = 2x + 8 \) and set it to zero: \[ 2x + 8 = 0 \quad \Rightarrow \quad x = -4. \] This value \( x = -4 \) gives us the minimum point of the quadratic function, and substituting it back into \( g(x) \) will give us the minimum value. Now for some extra fun facts! Did you know that parabolas, like the one formed by the quadratic function \( g(x) \), have a unique vertex that is either a maximum or a minimum point? For this function, we’ve just found that minimum! Also, when you're looking for the minimum value of a quadratic equation in the form \( ax^2 + bx + c \), you can always find the vertex using the formula \( x = -\frac{b}{2a} \). In this case, with \( a = 1\) and \( b = 8\), you'll land right back at \( x = -4 \). Math magic!