Expert: Ahmed Salami - 1/20/2010

Question
find the equation of the tangent and the normal to the curve at the point indicated:f(x)=3x^2-2x+1 @ (1,2)

Answer
Hi Zyryl,
The function f(x) = 3x² - 2x + 1 has a slope function of
f'(x) = 6x - 2
at (1,2), slope is
f'(1) = 6(1) - 2
     = 4
The tangent line at this point also has a slope of 4 and so its equation is
(y-2)/(x-1) = 4
y - 2 = 4(x-1)
y - 2 = 4x - 4
y = 4x - 2
The normal line has a slope equal to -1/m where m is the slope of the tangent line. The slope is therefore -1/4 and its equation is
(y-2)/(x-1) = -1/4
(y-2) = -(x-1)/4
y - 2 = (1 - x)/4
y - 2 = 1/4 - x/4
y = 9/4 - x/4
y = (9 - x)/4

Regards