for the function (f(x)=6x^{2}+10x + 7), find the slope of the tangent line at (x=-4).

Explanation:

Step1: Recall derivative formula

The derivative of a function $y = ax^{2}+bx + c$ is $y'=2ax + b$. For $f(x)=6x^{2}+10x + 7$, its derivative $f'(x)=12x + 10$.

Step2: Evaluate derivative at $x=-4$

Substitute $x = - 4$ into $f'(x)$. So $f'(-4)=12\times(-4)+10$. $f'(-4)=-48 + 10=-38$.

Answer:

$-38$