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

Apr 23, 2018

See below:

Explanation:

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

$f \left(x\right) = 6 x - {x}^{2}$

$f ' \left(x\right) = 6 - 2 x$

Hence:

$f ' \left(- 1\right) = 6 + 2 = 8$

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
$y = 8 x$

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 $f$.

$f \left(- 1\right) = - 6 - \left(1\right) = - 7$

So a point on the tangent line is $\left(- 1 , - 7\right)$

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

gradient$= \frac{\Delta y}{\Delta x}$

Hence:

$\frac{y - \left(- 7\right)}{x - \left(- 1\right)} = 8$

$y + 7 = 8 x + 8$

$y = 8 x + 1$

Apr 23, 2018

$\implies f \left(x\right) = 8 x + 1$

Explanation:

We are given

$f \left(x\right) = 6 x - {x}^{2}$

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.

$f ' \left(x\right) = 6 - 2 x$

Substituting our point $x = - 1$

$f ' \left(- 1\right) = 6 - 2 \left(- 1\right) = 6 + 2 = \textcolor{b l u e}{8}$

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

$f \left(- 1\right) = 6 \left(- 1\right) - {\left(- 1\right)}^{2} = - 6 - 1 = - 7$

So the point on the line is $\textcolor{b l u e}{\left(- 1 , - 7\right)}$.

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

$y - {y}_{p} = m \left(x - {x}_{p}\right)$

$y - \left(- 7\right) = 8 \left(x - \left(- 1\right)\right)$

$y + 7 = 8 x + 8$

$y = 8 x + 1$

Hence, the tangent line equation is: $\textcolor{b l u e}{f \left(x\right) = 8 x + 1}$

Apr 23, 2018

$y = 8 x + 1$

Explanation:

$\text{we require the slope m and a point "(x,y)" on the line}$

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

$\Rightarrow f ' \left(x\right) = 6 - 2 x$

$\Rightarrow f ' \left(- 1\right) = 6 + 2 = 8$

$\text{and } f \left(- 1\right) = - 6 - 1 = - 7 \Rightarrow \left(- 1 , - 7\right)$

$\Rightarrow y + 7 = 8 \left(x + 1\right)$

$\Rightarrow y = 8 x + 1 \leftarrow \textcolor{red}{\text{equation of tangent}}$