Question

Grade 12 calculus: derivatives?

What is the equation of the normal to y=(x²+3)/(x+1) at (1,2)?

Answer: x=1

I don't get why x=1 is the normal. I got 0/4 as my derivative for the equation, which means it's 0. Doesn't this mean that the normal of the equation is also 0 as well? :S

Answer 1

Finding the tangent line is easy. The slope of the line is the value of the derivative at the given point (1,2):\[y \prime=[2x(x+1)-(x^2+3)(1)]/(x+1)^2\]So,\[y \prime(1)=4-4/4=0/4=0\]as you said. This is the slope of the tangent line...not the normal line. The normal line is perpendicular to the tangent line. Our tangent is horizontal (it is y=2 actually) so we know that our normal line will be vertical (i.e., will be x=something). We also know that our normal line passes through the given point (1,2). Thus the normal line must be x=1.

Answer 2

as after differentiating y at at (1,2) we get 0 this means our tangent has slope 0. As normal is perpendicular to tangent hence slope of normal is (-1/0). because two perpendicular line's slope product is -1.
hence our normal is passing through (1,2) and has slope (-1/0).
eq.. of normal is y-2=(-1/0)(x-1)
-> 0(y-2)=-1(x-1) cross multiplication
-> 0=-1(x-1)
-> 0=x-1
hence x=1 is our normal line.

Answer 3

Thanks guys!