Tangent Line/Normal Line?

Question

The problem says the find the equations for both the tangent line and the normal line to the curve
y=(x^2+3)/(2x) at the point (1,2).

Answer 1

y = (1/2) x + (3/2) x ^(-1)
dy/dx = (1/2) - (3/2) x^(-2)
dy/dx = (1/2) - 3 / 2x²
dy/dx = 1/2 - 3/2 at x = 1
dy/dx = - 1 = gradient of tangent line at x = 1
Equation of tangent line is:-
y - 2 = (-1) (x - 1)
y = - x + 3

Equation of normal :-
gradient is 1
y - 2 = (1) (x - 1)
y = x + 1

Answer 2

I will get the tangent as I don't understand what do u mean by normal line

y=mx+b
to get m get f' at this point
f(x)=(x^2+3)/(2x) = x/2 + 3/2x
f'(x)=1 - 3/(2x^2)
f'(1)=1-3/2= - 1/2
so m= -1/2
y= -1/2 x +b
to get b use the point (1,2)
1 = (-1/2 )*2 +b
1=-1+b
b=2
so the line equation of the tangent is
y= -x/2 +2

Answer 3

Take the derivative.
Plug the value x=1 into the derivative and you will get the slope of the tangent line.
Use the point-slope formula to get the equation for the tangent line.
Take the negative reciprocal of the tangent's slope, that's the slope of the normal line.
Finally, use the poinit-slope form again to get the equation for the normal line.

hope it helps!

Answer 4

y=(x^2+3)/(2x) at the point (1,2).

y= 2x-6x^-1
f'(x) = 2x^-1 - 6x^-2
f'(x) = 1/ (2x-6x^2)

Point (1,2)
therefore the gradient will be: -1/4 (by substituting x value into the equation)

y = -1/4x + c
2 = -1/4 + c
c = 2 + 1/4
c = 9/4
y = -1/4x + 9/4 or y = -0.25x + 2.25

for normal:
gradient: -1/-1/4 = 4
y = 4x +c
2 = 4 + c
c = -2
y = 4x -2

Answer 5

---Derivate:
---y' = 2x . 2x / 4x² - 2 (x² + 3) / 4x²
---y' = 4x² / 4x² - 2 (x² + 3) / 4x²
---y' = 1 - (x² + 3) / 2x² -> tangent line!
---for x = 1 then -> y' = 1 - (1² + 3) / 2x1² -> y' = - 1.