Question

Solve only number 6 ​

Answer 1

To find the tangent line of the graph:

⇒ must find:

• point at which the tangent line touches the graph
• slope of the tangent line

Let's find the point at which the tangent line touches the graph:

At x = -1,

`y = ((-1)^3+2)^5=(-1+2)^5=1^5=1`

Point: (-1,1)

Let's find the slope of the tangent line

⇒ get the derivation of function than plug (-1) in the x-position to get

the exact slope

`(d)/(dx)(x^3+2)^5=5(x^3+2)^4*(d)/(dx)(x^3+2)=5(x^3+2)^4*3x^2\\ \\ 5((-1)^3)+2)^4*3(-1)^2=5(-1+2)^4*3*1=5*(1)^4*3=5*1*3=15`

Slope of tangent line: 15

Now put all the calculated value into the point-slope form:

`(y-y_(0))=m(x-x_(0) )`

• 
  `(x_(0) ,y_(0) )` --> point on the tangent line
• m --> tangent slope's value

So:

`(y-1)=15(x+1)\\y - 1= 15x + 15\\y = 15x + 15 + 1\\y = 15x + 16`

Thus the tangent line's equation is y = 15x + 16

Answer: y = 15x + 16

Hope that helps!