Question

How do you find the equation of a line tangent to the function `y=x^2-2` at x=1?

Answer 1

`y=2x-3`

Explanation:

The equation of the tangent in `color(blue)"point-slope form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))`
where m represents the slope and `(x_1,y_1)" a point on the line"`

Now `m=dy/dx` and substituting x = 1 into the function will give us a point on the line.

`y=x^2-2rArrdy/dx=2x`

`x=1rArrdy/dx=2xx1=2=m`

`x=1rArry=(1)^2-2=-1`

substitute m = 2 and `(x_1,y_1)=(1,-1)" into the equation"`

`rArry-(-1)=2(x-1)rArry+1=2x-2`

`rArry=2x-3" is the equation"`