Question

How do you find an equation of the tangent line to the graph `y=x^4-3x^2+2` at (1,0)?

Answer 1

`y=-2x+2`

Explanation:

The tangent line has the same slope as the equation at the point of contact.

To find the slope of the tangent line, we find the slope of the equation at `(1,0)`. For this, we use calculus.

`(dy)/dx=(d(x^4-3x^2+2))/dx=4x^3-6x`

At `(1,0)`, the slope is `4*1^3-6*1=-2`. The point-slope form of a linear equation is `y-y_0=m(x-x_0)`, where `(x_0,y_0)` is an arbitrary point on the line and `m` is the slope.

We know the slope is `-2`, and that `(1,0)` is a point on a line. Substituting these values in, we get `y-0=-2(x-1)`. Simplifying, we get `y=-2x+2`.