# `y = x^4 + 2x^2 - x, (1, 2)` Find an equation of the tangent line to the curve at the given point.

*print*Print*list*Cite

### 2 Answers

**NOTE: 1) If f(x) = x^n ; where n = integer; then **

**f'(x) = d[f(x)]/dx = n*x^(n-1) **

**2) If f(x) = k ; where 'k' = real number, then**

**f'(x) = d[f(x)]/dx = 0**

Equation of curve :-

y = x^4 + 2*(x^2) - x

Differentiating the given fuction with respect to x we get

y'(x) = 4*(x^3) + 4*(x^1) - 1*(x^0) = 4*(x^3) + 4x -1

Now, y'(x) at (1,2) = slope of the tangent to the curve at (1,2)

or, y(1) = 4 + 4 - 1 = 7

**Now, equation of line passing through a point (a,b) and having a slope 's' is given by the equation :-**

**y-b = s*(x-a)**

Thus, equation of the tangent to the curve at (1,2) is :-

y-2 = 7*(x-1)

or, y-2 = 7x - 7

or, y - 7x + 5 = 0 is the equation of the tangent to the given curve.

Given curve is y=x^4+2x^2-x

The slope of the tangent at a point is equal to the derivative of the curve at that point.

`(dy)/(dx)=4x^3+4x-1`

The slope of the tangent at (1,2) is 7

The equation of the tangent at (1,2) is y-2=7(x-1)

That is y=7x - 5.