Calculating tangent lines

Differentiation: Applications of derivatives

Calculating tangent lines

Before introducing the derivative, we saw that a tangent line is a line touching the graph at a point, meaning that the line's slope equals the graph's slope at that point. Once the derivative was introduced, we saw that the slope of a function at a point equals the derivative's function value at that point.

The tangent line $\orange l$ to the graph of $\blue f$ at the point $x=\green{p}$ is the line with slope $\purple{f'}(\green p)$ that passes through the point $\rv{\green p, \blue f(\green p)}$ .

Determining the tangent line is the same as determining any other line. Use the following step-by-step approach to finding the formula of a tangent line.

Determining tangent line

	Step-by-step	Example
	Determine the tangent line $\orange{l: y=a \cdot x+b}$ to the graph of $\blue f$ at the point $x=\green{p}$ .	$\qqquad \begin{array}{rcl}\blue{f(x)} & = & \blue{x^3-5 \cdot x^2+1} \\ x & = &\green{3}\end{array}$
Step 1	Calculate $\blue f(\green p)$ to find the $y$ -coordinate of the point through which the tangent line passes.	$\qqquad \begin{array}{rcl}\blue{f}(\green 3) &=& -17 \end{array}$
Step 2	The slope $\orange a$ of the tangent line is equal to $\purple{f'}(\green{p})$ . Determine the derivative $\purple{f^{\prime}}$ and substitute $x=\green p$ into $\purple{f'}(x)$ to evaluate $\purple{f'}(\green p)$ .	$\qqquad \begin{array}{rcl} \purple{f'(x)} &=& \purple{3 \cdot x^2-10 \cdot x} \\\purple{f'}(\green3) &=&\orange{-3} \end{array}$
Step 3	Line $\orange l$ has formula $\orange{y=} \purple{f'}(\green{ p})\orange{\cdot x+ b}$ . Determine $\orange b$ by substituting the point $\rv{\green p, \blue f(\green p)}$ into this formula.	$\qqquad \begin{array}{rcl}-17&\orange{=}&\orange{-3 \cdot} \green{3} \orange{+b}\\ \orange{b} &\orange{=}&\orange{-8}\end{array}$
Step 4	Substitute the value we found for $\orange b$ into the formula of the tangent line $\orange l$ .	$\qqquad \begin{array}{rcl}\orange {l: \ y} & \orange{=} & \orange{-3 \cdot x-8}\end{array}$

Find the formula for the tangent to the function $f(x)=-3\cdot x^3-4\cdot x+7$ at the point $x=-2$ .
Give your answer in the form $y=a\cdot x + b$ .

$y=-40\cdot x-41$

We follow the step-by-step plan for finding the formula of the tangent.

Step 1	We determine $f(-2)$ to know the $y$ -coordinate of the point through which the tangent line passes. $f(-2)=-3\cdot \left(-2\right)^3-4\cdot \left(-2\right)+7=39$ So the tangent goes through the point $\rv{-2, 39}$ .
Step 2	We calculate the slope of the tangent line. To do this, we first calculate $f'(x)$ . The function $f(x)=-3\cdot x^3-4\cdot x+7$ is a polynomial. We can calculate the derivative of a polynomial by using the sum rule and the derivative of the power function: $\dfrac{\dd}{\dd x} a x^n =a \cdot n x^{n-1}$ For the function $f(x)$ , we then find the derivative: $f^{\prime}(x) = -9\cdot x^2-4$ We now find the slope of the tangent line by substituting $x=-2$ into the derivative. $f'(-2)=-9\cdot \left(-2\right)^2-4=-40$
Step 3	We determine the $y$ -intercept $b$ of the formula of the tangent line. In step 1 we determined that the tangent line passes through the point $\rv{-2, 39}$ . We now substitute this point into the formula of the tangent line $y=-40\cdot x+b$ and solve the resulting equation $39 = -40\cdot \left(-2\right)+b$ for the unknown $b$ , giving $b=-41$
Step 4	Entering the calculated slope and $y$ -intercept gives the formula for the tangent line: $y=-40\cdot x-41$

The figure shows the function $f(x)=-3\cdot x^3-4\cdot x+7$ in blue, the point $P=\rv{-2, 39}$ in green, and the tangent to $f$ at the point $P$ , with formula $y=-40\cdot x-41$ , in orange.

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