# How do you find the slope of a tangent line to the graph of the function g(x) = 14 − x^2 at (2, 10)?

Jul 9, 2018

$\text{slope of tangent line } = - 4$

#### Explanation:

$\text{the slope of the tangent line } = f ' \left(2\right)$

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

$f ' \left(2\right) = - 2 \left(2\right) = - 4 \leftarrow \textcolor{b l u e}{\text{slope of tangent line}}$

Jul 9, 2018

Part 2 of 2

Using FIRST PRINCIPLES with explanation about determining the slope

#### Explanation:

$\textcolor{b l u e}{\text{The background bit}}$

Slope (gradient) is $\left(\text{change in "y)/("change in } x\right)$ reading left to right on the x-axis.

Suppose we have two points on the graph.

Let the first point be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right)$
Let the second point be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right)$

Then $\left(\text{change in "y)/("change in } x\right) \to \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

This you should have seen before!

In Calculus you start to bring the two points so close together that you can not physically measure the gap between them.

Suppose instead of using ${y}_{2} \mathmr{and} {x}_{2}$ we used ${y}_{1} + \delta y \mathmr{and} {x}_{1} + \delta x$ Where the symbol $\delta x$ means a minuscule bit of $x$

These days people use $\Delta y \mathmr{and} \Delta x$; I prefure the other notation.

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$\textcolor{b l u e}{\text{Using the above within the context of the question}}$

Set the fist point as ${P}_{1} \to \left(x , y\right)$
Let the next point progress by some minuscule value in $x + \delta x$. Consequently $y$ will also change by some minuscule amount $y + \delta y$

Set $y = - {x}^{2} + 14 \text{ } \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$

Then the progression point gives us:

$y + \delta y = - {\left(x + \delta x\right)}^{2} + 14$

$y + \delta y = - \left({x}^{2} + 2 x \delta x + {\left(\delta x\right)}^{2}\right) + 14$

$y + \delta y = - {x}^{2} - 2 x \delta x - {\left(\delta x\right)}^{2} + 14 \text{ } \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$
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$E q n \left(2\right) - E q n \left(1\right)$
$y + \delta y = - {x}^{2} - 2 x \delta x - {\left(\delta x\right)}^{2} + 14$
ul(ycolor(white)("dddd")=-x^2color(white)("dddddddddddd.")+14 larr" Subtract")
$\textcolor{w h i t e}{\text{ddd")deltay =color(white)("dd") 0color(white)("d}} - 2 x \delta x - {\left(\delta x\right)}^{2} + 0$

$\delta y = - 2 x \delta x - {\left(\delta x\right)}^{2}$

But we need the gradient which is $\left(\text{change in "y)/("change in } x\right)$

but change in $x$ is $\delta x$ so divide both sides by $\textcolor{red}{\delta x}$

$\textcolor{g r e e n}{\delta y = - 2 x \delta x - {\left(\delta x\right)}^{2} \textcolor{w h i t e}{\text{ddd")->color(white)("ddd}} \frac{\delta y}{\textcolor{red}{\delta x}} = 2 x \frac{\cancel{\delta x}}{\textcolor{red}{\cancel{\delta x}}} - {\left(\delta x\right)}^{\cancel{\textcolor{w h i t e}{.} 2}} / \textcolor{red}{\cancel{\delta x}}}$

$\frac{\delta y}{\delta x} = 2 x + \delta x$

We now make $\delta x$ so small that it so close to zero it may as well be 0.

${\lim}_{\delta x \to 0} \frac{\delta y}{\delta x} = \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x + 0$

$\textcolor{b l u e}{\overline{\underline{| \text{ The slope " -> dy/dx=g'(x)=2x" } |}}}$

Jul 9, 2018

Part 1 of 2
Using shortcut: $s l o p e \to - 4$

#### Explanation:

Given point $\left(x , y\right) = \left(2 , 10\right)$

Set $g \left(x\right) = y = - {x}^{2} + 14$

If we have $y = a {x}^{n} + c$ then $\frac{\mathrm{dy}}{\mathrm{dx}} = n a {x}^{n - 1}$

The constant of $c$ just 'goes away'. See solution part 2 of 2

So in this case:

The slop $g ' \left(x\right) = - 2 x = \left(\text{change in "y)/("change in } x\right) = \frac{- 2 x}{1}$

In other words it is the UNIT RATE OF CHANGE.

So at $\textcolor{red}{x = 2}$ the rate of change ->color(green)(g'(x)=-2color(red)(x) =-2(color(red)(2))=-4/1=("change in "y)/("change in "x)