find equation of tangent and normal lines (differentiation)
What is the equation of the tangent and normal lines to this function at the point p
$f(x)$ = $x^3$ at the point $p = (2,8)$
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$\begingroup$First, we take the derivative of $f(x)$
$f'(x) = 3x^2, \tag{1}$
the evaluate it at $x = 2$ to obtain the slope at $(2, 8)$:
$f'(2) = 3(2)^2 = 12. \tag{2}$
Knowing this, we have that the tangent line, that is, the line of slope $12$ passing through the point $(2, 8)$, is given by
$y - 8 = 12(x - 2), \tag{3}$
or
$y = 12x - 16. \tag{4}$
The normal line obeys a similar equation but its slope is $-\frac{1}{12}$; thus
$y - 8 = -\frac{1}{12}(x - 2), \tag{5}$
or
$y = -\frac{1}{12}x + \frac{49}{6} \tag{6}$
are equations for the normal line. QED.
Hope this helps. Cheerio,
and as alwaus,
Fiat Lux!!!
$\endgroup$ 3 $\begingroup$To find the slope of the needed lines, first find $f'(x)$ and evaluate at $x = 2$, since $p = (2, 8)$ is a point on both lines. The slope $m_1$ of the tangent line will equal $m_1 = f'(2)$. The slope $m_2$ of the normal line will equal $m_2 = -\dfrac 1{f'(2)} = -\dfrac 1{m_1}$.
For both lines you then have the slope, and the point on those lines: $(2, 8)$.
The point-slope form of an equation of a line with a given slope $m$ and which intersects the point $(x_0, y_0)$, is given by $$y - y_0 = m(x - x_0)$$
$\endgroup$ 1 $\begingroup$A line is determined by a slope and a point You have a point, $(2,8)$. Now obtain the slopes.
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