Finding Equations of Tangent Line

Consider a curve $y=f(x)$.

Finding Equations of Tangent Line

A tangent to a curve is a straight line which touches the curve at a given point and represents the gradient of the curve at that point.
If $A$ is the point with $x$-coordinate $a$, then the gradient of the tangent line to the curve at this point is $f'(a)$. The equation of the tangent is;
$$y-f(a) = f'(a)(x-a)$$

Example 1

Find the equation of the tangent line to $f(x)=x^2$ at the point where $x=3$.

\( \begin{align} \displaystyle \require{color}
f(3) &= 3^2 \\
&= 9 \\
f^{\prime}(x) &= 2x \\
f^{\prime}(3) &= 2 \times 3 \\
&= 6 \\
y-9 &= 6(x-3) &\color{red} y-f(3) = f^{\prime}(3)(x-3)\\
&= 6x-18 \\
\therefore y &= 6x-9 \\
\end{align} \)

Example 2

Find the equations of tangent lines to $f(x)=2x^2-8x$ at the point where the gradient is $4$?

\( \begin{align} \displaystyle \require{color}
f^{\prime}(x) &= 4 \\
4x-8 &= 4 \\
4x &= 12 \\
x &= 3 \\
f(3) &= 2 \times 3^2 – 8 \times 3 \\
&= -6 \\
y-(-6) &= 4(x-3) &\color{red} y-f(3) = 4(x-3) \\
\therefore y &= 4x-18 \\
\end{align} \)

Example 3

Find the equations of the tangents to the curve $f(x)=x^2+3x-10$ at the points where the curve cuts the $x$-axis.

\( \begin{align} \displaystyle \require{color}
f^{\prime}(x) &= 2x+3 \\
x^2+3x-10 &= 0 &\color{red} \text{the curve cuts the $x$axis} \\
(x+5)(x-2) &= 0 \\
x &= -5 \text{ or } x=2 \\
f^{\prime}(-5) &= 2 \times (-5)+3 &\color{red} \text{gradient at }x=-5\\
&= -7 \\
y – 0 &= -7(x–5) &\color{red} y-f(-5) = f^{\prime}(-5)(x–5)\\
y &= -7x-35 \\
f^{\prime}(-5) &= 2 \times 2+3 &\color{red}\text{gradient at }x=2\\
&= 7 \\
y – 0 &= 7(x-2) &\color{red} y-f(2) = f^{\prime}(2)(x-2)\\
y &= 7x-14 \\
\therefore y &= -7x-35 \text{ and } y = 7x-14 \\
\end{align} \)

Example 4

Find the equations of any horizontal tangent lines to $f(x)=2x^3-3x^2-12x+1$.

Finding Equations of Horizontal Tangent Line

$f^{\prime}(x) = 6x^2-6x-12$
Horizontal tangents have gradient $0$.
\( \begin{align} \displaystyle \require{color}
6x^2-6x-12 &= 0 \\
x^2-x-2 &= 0 \\
(x+1)(x-2) &= 0 \\
x &= -1 \text{ or } x=2 \\
f(-1) &= 2(-1)^3-3(-1)^2-12(-1)+1 \\
&= 8 \\
f(2) &= 2(2)^3-3(2)^2-12(2)+1 \\
&= -19 \\
\end{align} \)
The points of contact are $(-1,8)$ and $(2,-19)$.
Therefore the horizontal tangent lines are $y=8$ and $y=-19$.

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