The slope of the tangent line to the parabola y=3x^2+6x+5 at the point (-3,14)

To find the slope of the tangent line to the parabola at a given point, you need to find the derivative of the function at that point. The derivative of a function represents the rate at which the function is changing at a given point.

In this case, we want to find the derivative of the function y = 3x^2 + 6x + 5. To do this, we can use the power rule of differentiation. The power rule states that if we have a term ax^n, where a is a constant and n is a real number, then the derivative is given by d/dx(ax^n) = anx^(n-1).

Applying the power rule to each term in the function, we have:
- The derivative of 3x^2 is 2 * 3x^(2-1) = 6x.
- The derivative of 6x is 6.
- The derivative of 5 (a constant) is 0, since the derivative of a constant is always 0.

So, the derivative of y = 3x^2 + 6x + 5 is dy/dx = 6x + 6.

Now that we have the derivative, we can substitute the x-coordinate of the given point (-3, 14) into the derivative to find the slope of the tangent line at that point.

Substituting x = -3 into dy/dx = 6x + 6:
slope = dy/dx = 6(-3) + 6 = -18 + 6 = -12.

Therefore, the slope of the tangent line to the parabola y = 3x^2 + 6x + 5 at the point (-3, 14) is -12.