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Dmitry [639]
3 years ago
14

Y=x^2 -9x at x = 4 Find an equation of the tangent an an equation of the normal

Mathematics
1 answer:
Montano1993 [528]3 years ago
3 0

The tangent line to <em>y</em> = <em>f(x)</em> at a point (<em>a</em>, <em>f(a)</em> ) has slope d<em>y</em>/d<em>x</em> at <em>x</em> = <em>a</em>. So first compute the derivative:

<em>y</em> = <em>x</em>² - 9<em>x</em>   →   d<em>y</em>/d<em>x</em> = 2<em>x</em> - 9

When <em>x</em> = 4, the function takes on a value of

<em>y</em> = 4² - 9•4 = -20

and the derivative is

d<em>y</em>/d<em>x</em> (4) = 2•4 - 9 = -1

Then use the point-slope formula to get the equation of the tangent line:

<em>y</em> - (-20) = -1 (<em>x</em> - 4)

<em>y</em> + 20 = -<em>x</em> + 4

<em>y</em> = -<em>x</em> - 24

The normal line is perpendicular to the tangent, so its slope is -1/(-1) = 1. It passes through the same point, so its equation is

<em>y</em> - (-20) = 1 (<em>x</em> - 4)

<em>y</em> + 20 = <em>x</em> - 4

<em>y</em> = <em>x</em> - 24

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