Question

What is the equation of the tangent line of `f(x)=6x-x^2` at `x=-1`?

Answer 1

See below:

Explanation:

First step is finding the first derivative of `f`.

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

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

Hence:

`f'(-1)=6+2=8`

The value of 8's significance is that this is the gradient of `f` where `x=-1`. This is also the gradient of the tangent line that touches the graph of `f` at that point.

So our line function is currently
`y=8x`

However, we must also find the y-intercept, but to do this, we also need the y coordinate of the point where `x=-1`.

Plug `x=-1` into `f`.

`f(-1)=-6-(1)=-7`

So a point on the tangent line is `(-1,-7)`

Now, using the gradient formula, we can find the equation of the line:

gradient`=(Deltay)/(Deltax)`

Hence:

`(y-(-7))/(x-(-1))=8`

`y+7=8x+8`

`y=8x+1`

Answer 2

`=>f(x) = 8x+1`

Explanation:

We are given

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

To find the equation of the tangent line, we need to: find the slope of the tangent line, obtain a point on the line, and write the tangent line equation.

To find the slope of the tangent line, we take the derivative of our function.

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

Substituting our point `x = -1`

`f'(-1) = 6 - 2(-1) = 6+2= color(blue)(8)`

Now that we have our slope, we need to find a point on the line. We have an `x`-coordinate, but we need a `f(x)` too.

`f(-1) = 6(-1) - (-1)^2 = -6 - 1 = -7`

So the point on the line is `color(blue){(-1, -7)}`.

With a slope and a point on the line, we can solve for the equation of the line.

`y-y_p = m(x-x_p)`

`y - (-7) = 8(x - (-1))`

`y + 7 = 8x + 8`

`y = 8x + 1`

Hence, the tangent line equation is: `color(blue)(f(x) = 8x+1)`

Answer 3

`y=8x+1`

Explanation:

`"we require the slope m and a point "(x,y)" on the line"`

`•color(white)(x)m_(color(red)"tangent")=f'(-1)`

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

`rArrf'(-1)=6+2=8`

`"and "f(-1)=-6-1=-7rArr(-1,-7)`

`rArry+7=8(x+1)`

`rArry=8x+1larrcolor(red)"equation of tangent"`