need help really badly! please give answers in steps, much appreciated :)

Question

Find the equation of the tangent to the curve of y=2x^2+3x+1 at the point where x=1?

Answer 1

Equation of curve,y=2x^2+3x+1

At x=1

y=2(1)^2+3(1)+1

  =2+3+1

  =6

Slope of the tangent,m=dy/dx

Differentiate the equation of curve with respect to x

m=4x+3

At x=1

Slope is m=4(1)+3

               =7

Equation of the tangent,

(y-y1)=m(x-x1)

Substitute m=7,(x1,y1)=(1,6) in above equation

(y-6)=7(x-6)

Simpplify : y-6=7x-7

7x-y=1

Equation of the tangent is 7x-y=1