Answer :
Solution:
Given:
[tex]\begin{gathered} \text{The point (-2,3)} \\ f(x)=x^2-1 \end{gathered}[/tex]To get the slope using limit definition,
[tex]\begin{gathered} \frac{f(x+\Delta x)-f(x)_{}}{\Delta x} \\ \\ \text{Hence,} \\ \frac{f(x+\Delta x)-f(x)_{}}{\Delta x}=\frac{(x+\Delta x)^2-1-(x^2-1)}{\Delta x} \\ =\frac{x^2+2x\Delta x+\Delta x^2-1-x^2+1}{\Delta x} \\ =\frac{x^2-x^2+2x\Delta x+\Delta x^2-1+1}{\Delta x} \\ =\frac{2x\Delta x+\Delta x^2}{\Delta x} \\ =\frac{\Delta x(2x+\Delta x)}{\Delta x} \\ \frac{f(x+\Delta x)-f(x)_{}}{\Delta x}=2x+\Delta x \end{gathered}[/tex]Applying the limit,
[tex]\begin{gathered} \lim _{\Delta x\to0}\frac{f(x+\Delta x)-f(x)_{}}{\Delta x}=\lim _{\Delta x\to0}2x+\Delta x \\ =2x+0 \\ =2x \\ m=\lim _{\Delta x\to0}\frac{f(x+\Delta x)-f(x)_{}}{\Delta x}=2x \\ \\ \text{Hence, } \\ m=2x \end{gathered}[/tex]At the point (-2,3),
[tex]\begin{gathered} x=-2,y=3 \\ \text{Then the slope is;} \\ m=2x \\ m=2(-2) \\ m=-4 \end{gathered}[/tex]Using the equation of a line,
[tex]\begin{gathered} y=mx+b \\ \\ \text{Substituting the slope (m) into the equation,} \\ y=-4x+b \\ \\ To\text{ get the constant (b), substitute the point (-2,3)} \\ \text{where x =-2, y = 3} \\ y=-4x+b \\ 3=-4(-2)+b \\ 3=8+b \\ 3-8=b \\ -5=b \\ b=-5 \\ \\ \\ \text{Thus, the equation is;} \\ y=-4x+(-5) \\ y=-4x-5 \end{gathered}[/tex]Therefore, the equation of the tangent at (-2,3) is;
[tex]y=-4x-5[/tex]