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Equation of a tangent line

Equation of a tangent line

Given a curve defined by the equation $y = f(x)$, the equation of the tangent line at the point $P(x_0,y_0)$ can be calculated using the following process:
  1. Do you have the y-coordinate of the point $P$ $y_0 = f(x_0)$
  2. Do you have the derivative of $f(x)$?. Differentiate the curve, $\frac{dy}{dx} = \frac{d}{dx}f(x) = f'(x)$
  3. Evaluate the derivate at the point $P$ to give the gradient of the tangent line $m_t = f'(x_0)$
  4. we now have the information to use the line formula: $$ \begin{align} y - y_0 &= m_t(x-x_0)\\ y &= m_tx + (y_0- m_tx_0) \end{align}$$

The tangent line has a gradient of $m_t$ and a y-intercept of $y_0 - m_tx_0$

Equation of a tangent line 1

Determine the equation of the tangent line to the curve $y = 4 - x^2$ when $x = 1$
solution - press button to display
  1. When $x_0 = 1$, $y_0 = 4 - (1)^2 = 3$
  2. The derivative is $\frac{dy}{dx} = -2x$
  3. Evaluating the derivative at $x=1$ gives $m_t = -2$
  4. We can now use the equation of a line formula: $$ \begin{align} y - 3 &= -2(x - 1) \\ y &= -2x + 5 \end{align} $$

Equation of a tangent line 2

Determine the equation of the tangent to the curve $y = x^3 - 3x^2 - 24x$ when $x=4$ hence determine the other point of intersection between the tangent line and the curve.

solution - press button to display
  1. When $x = 4$, $y = -80$
  2. $\frac{dy}{dx} = 3x^2 - 6x - 24$
  3. $\left.\frac{dy}{dx}\right|_{x=4} = 0$
  4. $y + 80 = 0(x-4)$ hence the tangent line is $y = -80$

The other point of intersection satisfies the equation $x^3 - 3x^2 - 24x = -80$

This can be rewritten as a cubic equation, $x^3 - 3x^2 - 24x + 80 =0$. We also know that as $y = -80$ is tangent at $x=4$, then $x-4$ is a linear factor of the cubic equation. (In fact we know more, because y= -80 is a tangent, $(x-4)^2$ is a quadratic factor of the stated cubic equation)

The cubic equation can the be factorised as follows:

$$ \begin{align} x^3 - 3x^2 - 24x + 80 &= 0 \\ (x-4)(x^2+ x -20) &= 0 \\ (x-4)(x-4)(x+5) &= 0 \end{align} $$

The other solution is therefore $x = -5$