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

          $\frac{d}{dx}(x^3+2)^5=5(x^3+2)^4*\frac{d}{dx}(x^3+2)=5(x^3+2)^4*3x^2\\ \\ 5((-1)^3)+2)^4*3(-1)^2=5(-1+2)^4*3*1=5*(1)^4*3=5*1*3=15$

    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!