Calculus Homework Help!?

1) tangent line

general equation for a line is y - y1 = m(x - x1). substitute 2 for x1, 3 for y1, and 5 for m:

y - 3 = 5(x - 2)

y - 3 = 5x - 10

y = 5x - 7

2) normal line

The normal line and the tangent line are perpendicular. Perpendicular lines have slopes that are negative reciprocals of each other.

Therefore, m = -1/5.

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

y - 3 = -x/5 + 2/5

y = -x/5 + 17/5

You were asked for an equation of the TANGENT line. When the equation of the tangent line is evaluated at x = 2, y' = 5. Five is not the slope. it is the solution when x = 2.

Try something simple

y' = 2x + 1 + c...f'(2) does equal 5. Now the anti-derivative will give us f(x)

y or f(x) = x^2 + x + c.....f(2) = 3

2^2 + 2 + c = 3

6 + c = 3

c = -3

f(x) = x^2 + x -3.... f(2) does equal 3, both equation satisfy the original problem.

Check with your instructor. I'm sure he/she will tell you there are many possible solutions to this problem. If i am wrong, please email me and let me know so I can learn too.

You're given the point (2,3) and the slope m = 5. You should be able to easily construct a line using point-slope form.

y - y₁ = m(x - x₁)