I need help with a Calculus problem dealing with derivatives.?

Question

I seriously am stuck at this one problem right now. It is:

If f(2) = 3 and f'(2) = 5, find an equation of the tangent line, and the normal line to the graph of y = f(x) at the point where x = 2.

I know that there is an equation to use. I tried it, but I keep getting the wrong answer from the book. Can anyone please help me?

The answers my book gives are
y= 5x - 7 for the tangent line and
y = (-1/5)x + (17/5)

Answer 1

f'(2) = 5

Slope of tangent line = 5

y = 5x + c
Passes through (2, 3)
3 = 10 + c
c = -7
y = 5x - 7

The equation of the tangent line is:
y = 5x - 7.

Slope of normal * slope of tangent = -1
Slope of normal = -1/5

y = (-1/5)x + c
3 = -2/5 + c
15 = -2 + 5c
5c = 17

y = -x/5 + 17/5
5y = -x + 17
x + 5y = 17

Answer 2

The tangent line will have the equation y = a*x + b; a is the slope and equal to f'(x) at x = 2, so the tangent line equation is y = 2*x + b. At x = 2, the tangent line's y value is the same as f(x), so y = 3 at x = 2:

3 = 2*2 + b

3 = 4 + b

b = -1

the equation is then y = 2*x -1

The normal line intersects the same point, but the slope is -1/a, so it's equation is

y = -(1/2)*x + b

again, y must equal 3 at x = 2, so

3 = -1 + b

b = 4

and the normal lines equation is y = -(1/2)*x + 4

Answer 3

the derivative of function gives the slope of tangent

f'(x) = slope

the slope at x = 2 = f'(2)

but f'(2) = 5 , given

the slope, m = 5

the point of tangency = (x, f(x))

so here the point of tangeny = (2, f(2))

but f(2) = 3, given

so point of tangency = (2,3)

now we now the slope and point of tangency. So we can frame the equation of tangent

y - y1 = m(x - x1)

y - 3 = 5(x- 2)

y - 3 = 5x - 10

y - 5x + 7 = 0 is the equation of tangent.

The normal is perpendicular to tangent.

so the slope of normal = - 1/5

(since product of slopes of perpendicular lines = -1)

Normal also goes through (2, 3)

so equation of normal

y - 3 = -1/5(x - 2)

5y - 15 = -x + 2

5y + x = 17 is the equation of normal