Question

Find the equation of the tangent at (-2,3) using the limit definition.f(x) = x^2 - 1

Answer 1

Given:

`\begin{gathered} \text{The point (-2,3)} \\ f(x)=x^2-1 \end{gathered}`

To get the slope using limit definition,

`\begin{gathered} \frac{f(x+\Delta x)-f(x)_{}}{\Delta x} \\ \\ \text{Hence,} \\ \frac{f(x+\Delta x)-f(x)_{}}{\Delta x}=((x+\Delta x)^2-1-(x^2-1))/(\Delta x) \\ =(x^2+2x\Delta x+\Delta x^2-1-x^2+1)/(\Delta x) \\ =(x^2-x^2+2x\Delta x+\Delta x^2-1+1)/(\Delta x) \\ =(2x\Delta x+\Delta x^2)/(\Delta x) \\ =(\Delta x(2x+\Delta x))/(\Delta x) \\ \frac{f(x+\Delta x)-f(x)_{}}{\Delta x}=2x+\Delta x \end{gathered}`

Applying the limit,

`\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}`

At the point (-2,3),

`\begin{gathered} x=-2,y=3 \\ \text{Then the slope is;} \\ m=2x \\ m=2(-2) \\ m=-4 \end{gathered}`

Using the equation of a line,

`\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}`

Therefore, the equation of the tangent at (-2,3) is;

`y=-4x-5`