Expert: Paul Klarreich - 3/28/2006

Question
My name is Pernell. Studying calculus. I tried by taken the given: f(x)=2x^+ x, find the equation of the tangent line at the point of x=1/4.

I think the first derivative is 4x^,then I plugged 1/4 for every x.  2 1/4^1/2 + 1/4=2 1/2 + 1/4 = 2 3/4 now I am lost.  

Answer
Hi, Pernell,

You wrote:
Subject: first derivative to find slope of tangent line
Question:  My name is Pernell. Studying calculus. I tried by taken the given: f(x)=2x^+ x, find the equation of the tangent line at the point of x=1/4.

I think the first derivative is 4x^,then I plugged 1/4 for every x. 2 1/4^1/2 + 1/4=2 1/2 + 1/4 = 2 3/4 now I am lost.   
--------------------------------


Did you leave something out?  
Is the function  f(x) = x^2 + x?

If so, the derivative is f'(x) = 4x + 1,  not 4x.

Anyway, to find the equation of the T.L at x = 1/4, you have these steps to carry out:

1. Use your derivative to find the slope of the T.L.

 m = f'(1/4) = 4(1/4) + 1 = 1 + 1 = 2

2. Use the function to find the y-coordinate at x = 1/4:

 y = f(1/4) = 2(1/4)^2 + (1/4) = 2(1/16) + 1/4 = 1/8 + 1/4
   = 3/8

3. Now apply the point-slope form of the equation of a line:

y - y0 = m(x - x0)

Using:
m = 2
x0 = 1/4
y0 = 3/8

y - 3/8 = 2(x - 1/4)

Simplify a bit:

y - 3/8 = 2x - 1/2

Mult by the LCD = 8:

8y - 3 = 16x - 4
    1 = 16x - 8y

which should be OK.