# How do you find the equation of the tangent line to the curve y=x^4+2x^2-x at (1,2)?

Nov 11, 2016

$y = 7 x - 5$

#### Explanation:

We have; $y = {x}^{4} + 2 {x}^{2} - x$

First we differentiate wrt $x$;
$y = {x}^{4} + 2 {x}^{2} - x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} + 4 x - 1$

We now find the vale of the derivative at $\left(1 , 2\right)$ (and it always worth a quick check to see that $y = 2$ when $x = 1$) we have $\frac{\mathrm{dy}}{\mathrm{dx}} = 4 + 4 - 1 = 7$

So at the tangent passes through the coordinate $\left(1 , 2\right)$ and has gradient $m = 7$

We now use $y - {y}_{1} = m \left(x - {x}_{1}\right)$ to get the equation of the tangent:

$\therefore y - 2 = 7 \left(x - 1\right)$
$\therefore y - 2 = 7 x - 7$
$\therefore y = 7 x - 5$

Nov 11, 2016

The equation of the tangent is $y = 7 x - 5$

#### Explanation:

let $f \left(x\right) = {x}^{4} + 2 {x}^{2} - x$
Then the derivative is $f ' \left(x\right) = 4 {x}^{3} + 4 x - 1$

At the point $\left(1 , 2\right)$, $f ' \left(1\right) = 4 + 4 - 1 = 7$

So the slope of the tangent is $m = 7$
The equation of the line is, $y - 2 = 7 \left(x - 1\right)$
$y - 7 x = - 5$
graph{(y-x^4-2x^2+x)(y-7x+5)=0 [-5.55, 5.55, -2.773, 2.776]}