Question

Find an equation of the tangent line?

to the curve y= −5−2x−3x^2 at (1,−10).

Answer 1

see below....

Explanation:

`y=-5-2x-3x^2`

`=>(dy)/(dx)=d/(dx)(-5-2x-3x^2)`

`=>(dy)/(dx)=-2-6x" "`[Using the rule `color(red)(d/(dx)x^n=n cdot x^(n-1)`]

`=>(dy)/(dx)=-2-6 cdot 1=-8`

• The gradient of the tangent is `-8`

The equation of line is

`(y+10)/(x-1)=-8`
`=>y+10=-8x+8`
`=>color(red)(ul(bar(|color(green)(8x+y+2=0)|`

Answer 2

`color(blue)(y=-8x-2)`

Explanation:

To find the equation of the tangent line to a given point, we first find the first derivative. This will allow us to find the gradient of the tangent line at the given point.

`f(x) -5-2x-3x^2`

`f'(x)=-2-6x`

Plugging in `x=1`

`-2-6(1)=-8`

So gradient `bbm=-8`

Using point slope form of a line:

`(y_2-y_1)=m(x_2-x_1)`

We have a point on the line `(1,-10)`

Plugging in these values and the gradient value:

`y-(-10)=-8(x-(1))`

`color(blue)(y=-8x-2)`

This is the equation of the tangent line to the point `(1,-10)`

The graph confirms this result: