# Math Help - Tangent lines and derivatives

1. ## Tangent lines and derivatives

"Which of the following is an equation of the line tangent to the graph of
$f(x)= x^4 + 2x^2$ at the point where $f'(x)=1$? "

I found the derivative/slope to be:
$f'(x)= 4x^3 + 4x$
And since f'(x) = 1, I set it equal to one.

$1=4x^3 + 4x$

Even if I was able to factor that out (which I can't figure out how) I don't know how to find the equation of the tangent line without a point to work from.

2. Suppose you have factored it and x comes out to be some x'. Then what you want is a line that passes through (x', f(x')) with slope 1. Clearly, this line is unique. Consider the usual form of the equation of a line: y = mx + c. Here m is the slope and c is the intercept of the line with the y axis. You already have m=1, x=x', y=f(x'). So you can solve for c to get the equation.

3. Originally Posted by Traveller
Suppose you have factored it and x comes out to be some x'. Then what you want is a line that passes through (x', f(x')) with slope 1. Clearly, this line is unique. Consider the usual form of the equation of a line: y = mx + c. Here m is the slope and c is the intercept of the line with the y axis. You already have m=1, x=x', y=f(x'). So you can solve for c to get the equation.
m=1
$y-f(x')= 1(x - x')$
$y = x- x' + f(x')$
So $c= x' + f(x')$ ?

a. y = 8x - 5
b. y= x + 7
c. y= x + .763
d. y= x - .122
e. y= x - 2.146

It can't be a. because m=1.
But using the notation you gave me, all I get is a bunch of variables for c.

4. Your last step was wrong; c = f(x') - x'.

See if you can solve for x from the derivative by factorizing. If you can't do that, you can always use Cartan's formula. x will come out to be something around 0.23673290386456308 ( I got it from here : Cubic equation Calculator - Calculation ). Find y and see which answer holds.