How to find an equation fo the tangent line and normal line to the graph of the equation at the point

Function: y=(x+3)^3
Point: (-2, 1)

I am so confused on what to do

Differentiating the function will give you the gradient at the point.

To make life easier for me I'm going to expand the function: (just the way my brain works)

y = (x+3)(x+3)(x+3) = (x+3)(x^2+6x+9) = x^3 + 6x^2 +9x + 3x^2 + 18x + 27

Gather the x^2 and x terms:

y = x^3 + 9x^2 + 27x +27

Now differentiate:

dy/dx = 3x^2 +18x + 27

Substitute in the x value for your point, and you will have the slope of the tangent line at that point:

m = 3(-2)^2 + 18(-2) + 27 = 12 - 36 +27 = 3

Now that you know the slope you can substitute in the y value of your point to find the intercept:

y = mx + c --> 1 = 3(-2) + c --> c = 7

So the equation of the tangent line is just y = 3x + 7

To find the normal line, remember that the line at right angles to another line has a slope of -1/m where m is the slope of the original line. So the slope of the normal line will be -1/3.

You can substitute in the x and y values of your point again to find the intercept of the normal line:

y = mx + c --> 1 = -1/3(-2) + c ---> c = 1/3

So the equation of the normal line is y = -1/3x + 1/3