Originally Posted by

**mark** hi, i've tried to do this question but my answer isn't coming out the same as my textbooks. just seeing if someone can show me what i'm doing wrong

find an equation for the normal to the following curve at the point where x has the given value:

$\displaystyle y = (x^3 - 6)(2 - 7x)$ and x = -1

first i expanded the brackets and got the equation $\displaystyle 2x^3 - 7x^4 - 12 + 42x$

then used x = -1 and came up with - 2 - 7 - 12 - 42 which is - 63. so y = -63

then i differentiated the equation $\displaystyle \frac{dy}{dx}= 6x^2 - 28x^3 + 42$ and used x = -1 to come up with the gradient 76

6(-1)^2 - 28(-1)^3 + 42 = 6 +

i know the gradient of the norm will be perpendicular to the tangent so the gradient of the norm will be $\displaystyle -\frac{1}{76}$

so the equation of the norm will be $\displaystyle y + 63 = -\frac{1}{76}(x + 1)$

which comes to $\displaystyle y = -\frac {1}{76}x - \frac{1}{76} - \frac{4788}{76}$ and then $\displaystyle 76y = - x - 4789$

the answer my book gives is 8y + 503 = x

any help would be appreciated

thanks, mark