What is the equation of the line that is normal to $f (x) = {(x + 4)}^{2} - e^{x}$ $f(x)=(x+4)^2-e^x$ at $x = - 3$ $x=-3$ ?

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Answer 1 · 2018-06-10T00:02:56

$y = \frac{e^{3} x}{1 - 2 e^{3}} + \frac{3 e^{3}}{1 - 2 e^{3}} + \frac{e^{3} - 1}{e^{3}}$ $y = (e^3x)/(1-2e^3) +(3e^3)/(1-2e^3) +(e^3 - 1)/e^3$

or $y = \frac{e^{3} x}{1 - 2 e^{3}} + \frac{e^{6} + 3 e^{3} - 1}{e^{3} (1 - 2 e^{3})}$ $" "y = (e^3x)/(1-2e^3) + (e^6 + 3e^3 - 1)/(e^3(1-2e^3)$

Given: $f (x) = {(x + 4)}^{2} - e^{x} at x = - 3$ $f(x) = (x + 4)^2 - e^x " at "x = -3$

First find the slope of the curve at $x = - 3$ $x = -3$ .

The first derivative is called the slope function. We need to find the first derivative:

$f'(x) = 2(x+4)^1 (1) - e^x$

$f'(x) = 2x + 8 - e^x$

slope = $m = f'(-3) = 2(-3) + 8 - e^-3 = 2 - 1/e^3 = (2e^3 - 1)/e^3$

A line that is normal is a perpendicular line. $m_("perpendicular") = -1/m: " "-e^3/(2e^3 - 1)$

$f(-3) = (-3 + 4)^2 - e^-3 = 1 - 1/e^3 = (e^3 - 1)/e^3$

To find the Normal line, use the point $(-3, (e^3 - 1)/e^3)$ :

$y - y_1 = m (x - x_1)$

$y - (e^3 - 1)/e^3 = -e^3/(2e^3 - 1)(x +3)$

$y - (e^3 - 1)/e^3 = -e^3/(2e^3 - 1) x -(3e^3)/(2e^3 - 1)$

$y = -e^3/(2e^3 - 1) x -(3e^3)/(2e^3 - 1) +(e^3 - 1)/e^3$

$y = (e^3x)/(1-2e^3) +(3e^3)/(1-2e^3) +(e^3 - 1)/e^3$

Find a common denominator for the last two terms:

$y = (e^3x)/(1-2e^3) + (e^6 + 3e^3 - 1)/(e^3(1-2e^3)$

What is the equation of the line that is normal to f(x)=(x+4)^2-e^xf(x)=(x+4)2−exf(x)=(x+4)^2-e^x at x=-3 x=−3 x=-3 ?