What is the equation of the line normal to $f (x) = 2 x^{2} - x - 1$ $f(x)=2x^2-x - 1$ at $x = 5$ $x=5$ ?

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Answer 1 · 2015-12-25T00:46:28

$y = - \frac{1}{19} x + 44 \frac{5}{19}$ $y = -1/19x +44 5/19$

or

#y = -1/19 x + 841/19

Remember $m = f'(x)$ is slope of tangent line

Slope of normal line is $m_2 = 1/m$ as it is perpendicular to the tangent line

$f(x) = 2x^2 - x - 1$ at $x= 5$

Step 1: Find derivative, to determine the slope of tangent line

$f'(x) = d/dx(2x^2) -d/dx(x) -d/dx(1)$

$f'(x) = 4x -1$

Step 2: Find the slope of the tangent line

$m= f'(5) = 4(5) -1 = 19$

Slope of normal line is $m_2 = (-1)/19$

Step 3 Determine the $y$ coordinate, of $f(x)$ , when $x= 5$

$f(5) = 2(5)^2 -5 -1$

$f(5) = 50-5-1$

$f(5) = 44$

Ordered pair $(5, 44)$

Step 4: Write the equation of the line using point slope formula

$m_2 = -1/19 , " " " (5, 44)$

$y -44 = -1/19(x-5)$

$y-44 = -1/19 x +5/19$

$y = -1/19x + 5/19 +44$

$y = -1/19x +44 5/19$

or
$y = -1/19 x + 841/19$

What is the equation of the line normal to f(x)=2x^2-x - 1f(x)=2x2−x−1f(x)=2x^2-x - 1 at x=5x=5x=5?