Question

Find an equation of the normal to the curve at the point where x= -2, givingthe answer in the formax + by + C = 0, where a, b and c are integers.

Answer 1

`y=(x^3+7)^4(x^2-3)^3`

Let's find the derivative of this curve:

`(dy)/(dx)=12x^2(x^3+7)^3(x^2-3)^3+6x(x^2-3)^2(x^3+7)^4`

Evaluate it for x = -2:

`(dy)/(dx)\begin{cases} \\ x=-2\end{cases}=12(4)(-1)(1)+6(-2)(1)(1)=-48-12=-60`

Let:

m1 = -60

Since the line is normal, the other line, must have a slope m2 of the form:

`\begin{gathered} m1\cdot m2=-1 \\ m2=-(1)/(m1) \\ m2=-(1)/(-60) \\ m2=(1)/(60) \end{gathered}`

Using the point-slope equation:

`\begin{gathered} y-y1=m(x-x1) \\ _{\text{ }}where\colon \\ y1=y(-2)=(1)(1)=1 \\ so\colon \\ y-1=(1)/(60)(x+2) \\ y=(1)/(60)x+(31)/(30) \end{gathered}`

Rewrite the equation as: Ax + By + C = 0

`\begin{gathered} (1)/(60)x-y+(31)/(30)=0 \\ x-60y+62=0 \end{gathered}`

Answer:

x - 60y + 62 = 0