Given \( h(x)=x^{3}+6 \), find the equation of the secant line passing through \( (-4, h(-4)) \) and \( (2, h(2)) \). Write your answer in the form \( y=m x+b \).

First, let's calculate \( h(-4) \) and \( h(2) \): \[ h(-4) = (-4)^3 + 6 = -64 + 6 = -58 \] \[ h(2) = (2)^3 + 6 = 8 + 6 = 14 \] Now we have two points: \( (-4, -58) \) and \( (2, 14) \). Next, we find the slope \( m \) of the secant line, which is calculated using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Substituting in our points \( (-4, -58) \) and \( (2, 14) \): \[ m = \frac{14 - (-58)}{2 - (-4)} = \frac{14 + 58}{2 + 4} = \frac{72}{6} = 12 \] Now that we have the slope \( m = 12 \), we can use the point-slope form of the equation of a line, which is \( y - y_1 = m(x - x_1) \). We'll use the point \( (2, 14) \): \[ y - 14 = 12(x - 2) \] Expanding this equation: \[ y - 14 = 12x - 24 \] \[ y = 12x - 10 \] Thus, the equation of the secant line in \( y = mx + b \) form is: \[ \boxed{y = 12x - 10} \]