14. [-/2 Points] DETAILS MY NOTES HARMATHAP12 9.4.017. For the function \( P(x)=x^{3}-6 x \), at the point \( (2,-4) \), find the following. (a) the slope of the tangent to the curve (b) the instantaneous rate of change of the futtion Need Help? Readit SUBMIT ANSWER

To find the slope of the tangent to the curve at the point \( (2, -4) \), you first need to compute the derivative of the function \( P(x) = x^3 - 6x \). The derivative, \( P'(x) \), is \( 3x^2 - 6 \). Plugging in \( x = 2 \) gives us \( P'(2) = 3(2^2) - 6 = 12 - 6 = 6 \). So, the slope of the tangent at that point is \( 6 \). The instantaneous rate of change of the function at any point is simply the value of the derivative at that point. Since we already found that \( P'(2) = 6 \), it indicates that the instantaneous rate of change of the function at \( (2, -4) \) is also \( 6 \). This means for every unit you move along the x-axis at \( x = 2 \), the function value changes six units in the y-direction!