Tangent lines to the curve

• Jun 2nd 2010, 06:23 PM
AlanaSan
Tangent lines to the curve
Hello,

I need to find out the tangent lines to the curve f (X) = 2X² - 8X + 3,

Which are parallels to line X + 2Y + 1 = 0.

I have slope of line : m = -1/2,

and derivative : f'X = 4X - 8,

I made : m = f'X,

and I obteined : X = 1 7/8,

I replace in : f(X) and reach to Y = -4 31/32,

With those values I get Y intersection = -129/32.

Then I built my equation,
but I think something is wrong, cause those are a very big fractions.
But I cant find the mistake. Maybe someone could show me the error in this operation?
• Jun 2nd 2010, 07:48 PM
chiph588@
• Jun 2nd 2010, 07:51 PM
sa-ri-ga-ma
The point on the curve where the tangent is drawn is [tex](\frac{15}{8}, -\frac{1589}{32})/MATH]

Equation of the line parallel to x + 2y + 1 = 0 is x + 2y + c = 0.

Substitute the above point in the equation and find c.
• Jun 2nd 2010, 07:52 PM
sa-ri-ga-ma
The point on the curve where the tangent is drawn is (15/8, -159/32})

Equation of the line parallel to x + 2y + 1 = 0 is x + 2y + c = 0.

Substitute the above point in the equation and find c.