Answer :
[tex]\begin{gathered} f(x)=4x^2-2x+3 \\ \frac{d}{dx}f(x)=\frac{d}{dx}(4x^2-2x+3) \end{gathered}[/tex]
Applying the sum and difference rules:
[tex]\frac{d}{dx}f(x)=\frac{d}{dx}(4x^2)-\frac{d}{dx}(2x)+\frac{d}{dx}(3)[/tex]Applying the constant multiple rule:
[tex]\frac{d}{dx}f(x)=4\frac{d}{dx}(x^2)-2\frac{d}{dx}(x)+\frac{d}{dx}(3)[/tex]The derivative of a constant is zero, and the derivative of x is one. Applying the power rule:
[tex]\begin{gathered} \frac{d}{dx}f(x)=4(2x^{})-2(1)+0 \\ f^{\prime}(x)=8x^{}-2 \end{gathered}[/tex]Evaluating f'(x) at x = -4:
[tex]\begin{gathered} f^{\prime}(-4)=8(-4)^{}-2 \\ f^{\prime}(-4)=-34 \end{gathered}[/tex]This value is the slope of the tangent line at the point (-4, 75), that is,
[tex]m=-34[/tex]Given the general equation of a line:
[tex]y=mx+b[/tex]Substituting with m = -34 and the point (-4, 75), that is, x = -4, and y = 75, and solving for b:
[tex]\begin{gathered} 75=(-34)(-4)+b \\ 75=136+b \\ 75-136=b \\ -61=b \end{gathered}[/tex]And the equation of the tangent line is:
[tex]y=-34x-61[/tex]