how to solve this question?

Question

find the equation of the tangent to the curve y=x^2-x-1 at the point A(2,1). calculate the coordinates of the point on the curve at which the normal is parallel to the tangent at A.

Answer 1

f `(x) = 2x - 1
f `(2) = 4 -1 = 3 = gradient at (2,1)
Equation of tangent at (2,1) is given by:-
y - 1 = 3(x - 2)
y - 1 = 3x - 6
y = 3x - 5 is equation of tangent at A(2,1)

The reqd. normal is parallel to tangent at (2,1) therefore gradient of normal is 3

Gradient of curve at x is given by f `(x) = 2x - 1
Thus 3 x (2x - 1) = -1
6x - 3 = -1
6x = 2
x = 1/3

y = (1/3)² - (1/3) - 1 = (1/9) - 3/9 - 9/9 = -11/9
Reqd. co-ordinates are (1/3,-11/9)

Answer 2

the slope of the tangent is the derivade so slope =m =2x-1.
At(2,1) m= 2*2-1=3
the equ. is (y-1)= 3(x-2)==> y= 3x-5
At any point of a curve the normal is perpendicular at this point to the tangent.
If you have a point (a,b) b=a^2-a+1.The tangente is
(y-b)=(2a-1)(x-a) and the normal y-b= 1/(1-2a)(x-a)

The slope of the normal is 1/(1-2a) =3 ( for beeing paralel to the tangente at A
1= 3-6a ===> a= 1/3.Going to the curve b= 1/9-1/3-1 =-11/9

The point is (1/3,-11/9)

Answer 3

I don't know what the 2nd part means. but for the 1st part here is what you do:

To find the tangest line:

1) Derive the original equation to get to the slope of the equation:
in our case: the deriviation of y = x^2 - x - 1 is y' = 2x - 1.

2) To find the slope we know we are looking at point (2,1) so just plug in the x value into the slope equation: y' = 2*2 - 1 or y' = 3. So our slope is 3.

3) We need to find the equation of the tangent line so we use the formula (y - y1) = m*(x - x1) where x1,y1 is the point you are at.

so we get (y - 1) = 3*(x - 2)

taking the 3 though we get
y - 1 = 3x - 6

solve for y we get
y = 3x - 5 <--- thats your equation. i dunno what the 2nd part of the question means.

Answer 4

am doubt your gona get anyone who can answer that or who would want to answer maths??

soz