How do you find the equation of the line tangent to the graph of f(x) = x^4 + 2x^2?

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Answer 1 · 2015-08-03T22:46:15

Use differentiation and point slope equation to find equation of tangent at $(x_1, f(x_1))$ as:

$y = 4x_1(x_1^2+1)x-x_1^2(3x_1^2+2)$

If $f(x)$ is continuous and differentiable at $x_1$ , then the equation of the line tangent to $f(x)$ at $(x_1, f(x_1))$ in point slope form is:

$y - f(x_1) = f'(x_1)(x - x_1)$

Add $f(x_1)$ to both sides to get:

$y = f'(x_1)x + (f(x_1) - x_1f'(x))$

In our example, $f(x) = x^4+2x^2$ so

$f'(x) = 4x^3+4x = 4x(x^2+1)$

and the equation of the tangent in slope intercept form is:

$y = 4x_1(x_1^2+1)x + ((x_1^4+2x_1^2) - x_1(4x_1^3+4x_1))$

$=4x_1(x_1^2+1)x-x_1^2(3x_1^2+2)$