Find an equation of the tangent line to the graph y=(x^3 + 2)^5 at x= -1

dy/dx = 5(x^3 + 2)^4 (3x^2)

sub in x = -1 to find the slope of the tangent
sub x = -1 into original to get the corresponding y

Now that you have the slope and a point on the tangents, the equation is readily found

Let me know what you got

Find the value of y and dy/dx (the slope, m) at x' = -1. Call that value y'. Obviously y' = 1.

dy/dx is harder.

Then write the function in the form

m = (y - y')/(x - x')

To find the equation of the tangent line to a curve at a given point, you need to find the slope of the tangent line and the coordinates of the point of tangency. To do so, we can use the concept of derivative.

Step 1: Find the derivative of the function y=(x^3 + 2)^5 with respect to x.
To find the derivative, we can apply the chain rule. Let's denote f(x) = (x^3 + 2)^5. The chain rule states that the derivative of f(g(x)) is equal to f'(g(x)) * g'(x).
Using the chain rule, we have:
f'(x) = 5(x^3 + 2)^4 * (3x^2)
Simplifying, we get:
f'(x) = 15x^2(x^3 + 2)^4

Step 2: Calculate the slope of the tangent line at x = -1.
Substitute x = -1 into the derivative equation we found in step 1:
f'(-1) = 15(-1)^2((-1)^3 + 2)^4
Simplifying, we get:
f'(-1) = 15(1)(1 + 2)^4
f'(-1) = 15(1)(3)^4
f'(-1) = 15(1)(81)
f'(-1) = 1215

So, the slope of the tangent line to the graph at x = -1 is 1215.

Step 3: Find the coordinates of the point of tangency.
Substitute x = -1 into the original equation y = (x^3 + 2)^5:
y = (-1^3 + 2)^5
y = (1 + 2)^5
y = 3^5
y = 243

The coordinates of the point of tangency are (-1, 243).

Step 4: Write the equation of the tangent line.
We now have the slope (m = 1215) and a point on the line (x = -1, y = 243). We can use the point-slope form of the equation of a line to write the equation of the tangent line.
The point-slope form is given by:
y - y1 = m(x - x1)

Substituting the values we found, we get:
y - 243 = 1215(x - (-1))
y - 243 = 1215(x + 1)

Simplifying further, we get:
y - 243 = 1215x + 1215
y = 1215x + 1458

Therefore, the equation of the tangent line to the graph y = (x^3 + 2)^5 at x = -1 is y = 1215x + 1458.