Question

What is the equation of the tangent line of `f(x)=x^3+2x^2-3x+2` at `x=1`?

Answer 1

`y = 4x -2`

Explanation:

Step 1: Find derivative of the equation

`f(x) = x^3 + 2x^2 -3x +2`

`f'(x) = d/dx(x^3) + d/dx(2x^2) -d/dx(3x) + d/dx(2)`

`f'(x) = 3x^2 + 4x -3`

Step 2: Find the slope of the tangent line at `x = 1`
`m = f'(1) = 3(1^2) +4(1) -3 = 4`

Step 3: Find the `y` coordinate of the function when `x= 1`

`f(1) = (1)^3 + 2(1)^2 -3(1) +2`

`f(1) = 2`

So, the original point of the graph is `(1,2)`

Step 4: Find the equation of the tangent line using point slope formula

`y -y_0 = m(x-x_0)`

`m= 4 " " " " (1, 2)`

`y-2 = 4(x - 1)`
`=> y-2 = 4x - 4`
`=> y = 4x -4 +2`
`=> y = 4x -2`