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## Homework Statement

Determine the equations of the tangent lines to the graph of f(x)=3x(5x^2+1) that are parallel to the line y=8x+9

## Homework Equations

y=m(x-x_1 )+y_1

## The Attempt at a Solution

11.f(x)=3x(5x^2+1)

The slope of the tangent line y=8x+9

f^' (x)=3(5x^2+1)+3x(10x)

f^' (x)=15x^2+30x^2+3

f^' (x)=45x^2+3

Finding the point of tangency:

m=8

8=45x^2+3

5=45x^2

√(x^2 )=√(45/5)

x=±1/3

f(1/3)=(3(1/3))(5(1/3)^2+1)

f(1/3)=1(5(1/9)+1)

f(1/3)=5/9+9/9=14/9

f(1/3)=14/9

f(-1/3)=(3(-1/3))(5(-1/3)^2+1)

f(-1/3)=-1(5(1/9)+9/9)

f(-1/3)=-1(5/9+9/9)

f(-1/3)=-14/9

y=m(x-x_1 )+y_1

y_1=14/9 x_1=1/3 m=8

y=8(x-1/3)+14/9

y=8x-8/3+14/9

y=8x-24/9+14/9

y=8x-10/9

y=m(x-x_1 )+y_1

y_1=-14/9 x_1=-1/3 m=8

y=8(x-(-1/3))-14/9

y=8x+8/3-14/9

y=8x+24/9-14/9

y=8x+10/9

Therefore, the equations of the tangent lines are y=8x+10/9 and y=8x-10/9

I'm pretty sure I did this correctly, but i'm not confident in my final statement.