Question

please quickly help with some calculus problems

Answer 1

How do you find an equation of the normal line to the graph of y=x^3+3x^7+7x-1 at the point where x=1

Answer 2

Do you know how to start or anything?

Answer 3

@candy1975 The normal is perpendicular to the tangent at the point x = 1 of f(x). So first find the equation of the tangent line at x = 1. Then find the equation of the line perpendicular to this tangent line at x = 1 (i.e normal line). As you may already know, to find the perpendicular to the tangent at x = -1, you just need to find the negative reciprocal of the slope of the tangent, i.e. flip the slope of the tangent and negate, and that will be the equation of the normal.

So step 1: find the equation of the tangent at x = 1.
step 2: find the equation of the normal by flipping the slope and negating it.

@candy1975

Answer 4

Thx a bunch @genius12

Answer 5

@genius12 Please please please could you solve it and show me the steps cause i got confused on what to do after finding the derivative y'=3x^2+6x+7,

Answer 6

OK first find f'(x):\[\bf f'(x)=3x^2+6x+7\]Now we evaluate the derivative at x = 1 (this will give us the slope of the tangent line at at x = 1):\[\bf f'(1)=16\]Now to find the equation of the tangent line, we need a point that it goes through. We know the line goes through the point \(\bf (1,f(1))\) when \(\bf x = 1\). So let's now find f(1):\[\bf f(1)=1+3+7-1=10 \implies (1,f(1))=(1,10)\]Now we use this point and th slope for the equation of the tangent to get:\[\bf y = 16x+b \rightarrow 10=16(1)+b \implies b = -6 \implies y=16x-6\]So this is the equation of the tangent. Now the equation of the normally is found by changing the slope to its negative reciprocal. Can you do that? @candy1975

Answer 7

Yeah! totally get it now, thx so much :)