# Thread: Finding equation of a Tangent Line

1. ## Finding equation of a Tangent Line

Find an equation of the line that is tangent to the graph of f and parallel to the given line.

Function: f(x) = 2x^2

Line: 4xy + 4 = 0

Tell me what I'm doing wrong. According to my instructor's directions:

Step 1. Find the slope of the tangent line:

Do the algebra, reduces to 4x-2h - the limit/slope is 4x.

Step 2. Find the slope of the given line:

4x – y + 4 = 0 same as y=4x+4 and so m=4

Step 3. Set the slope of the tangent equal to the slope of the given line, and solve for x.

4x=4 so x=1

Step 4. Use the values from above to find the equation of the line.

(So elsewhere in the lecture, he mentions that if x=1, then your reference point is (1,1) - if someone else can explain why, I'd appreciate that too. I'm not getting that. But assuming that it is....)

y-1=4(x-1)

y-1=4x-4

y=4x-3

Why isn't this right? What am I missing? I'm sure it's something obvious and dumb?

2. $y - f(1) = 4(x - 1)$

$f(x) = 2x^2$

$f(1) = 2
$

$y - 2 = 4(x-1)$

$y = 4x - 2$

3. Nevermind. Naturally that "thing he said that made no sense to me" was the source of the problem, and I had to plug x back into the function to get y. Duh.

But thank you Skeeter!