Question

Find the equation of the line tangent to the graph of the function f(x)=2(x+1)^2 at the point (-2, 2).

Answer 1

Write out the function

`f(x)=2(x+1)^2`

To find the equation of the line tangent of this function, we different the function to obtain the slope.

Expanding the function we have

`\begin{gathered} f(x)=2(x+1)^2 \\ \text{expand the function } \\ f(x)=2\lbrack(x+1)(x+1)\rbrack \\ f(x)=2(x^2+2x+1) \\ \text{Hence} \\ f(x)=2x^2+4x+2 \end{gathered}`

Apply the differebtiation rule, we have

`\begin{gathered} f(x)=x^n,f^1(x)=nx^(n-1) \\ f(x)=2x^2+4x+2 \\ f^1(x)=4x+4 \\ \text{ Recall the derivative of a constant is zero} \end{gathered}`

hence, at the point (-2,2), the slope will be

`\begin{gathered} f^1(x)=4x+4 \\ \text{substitute x=-2} \\ \text{slope,m}=4(-2)+4=-8+4=-4 \\ \text{Hence} \\ \text{slope,m}=-4 \end{gathered}`

Then

Apply the Point- slope form for the equation of a line at point (-2,2)

`\begin{gathered} \text{The point slope form is given by } \\ y-y_1=m(x-x_1) \\ \text{Where } \\ m=-4,x_1=-2,y_1=2 \end{gathered}`

Substituting the value, we have

`\begin{gathered} y-2=-4(x-(-2)) \\ y-2=-4(x+2) \\ \\ \end{gathered}`

Exapand the paranthesis, and write the equation

`\begin{gathered} y-2=-4x-8 \\ \text{Add 2 to both sides } \\ y-2+2=-4x-8+2 \\ y=-4x-6 \end{gathered}`

Hence

The equation of the line of tangent to the function is

y=4x-6 or y+4x= -6