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2 years ago

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

1 answer:

Montano1993 [528]2 years ago

3 0

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

y = x² - 9x → dy/dx = 2x - 9

When x = 4, the function takes on a value of

y = 4² - 9•4 = -20

and the derivative is

dy/dx (4) = 2•4 - 9 = -1

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

y - (-20) = -1 (x - 4)

y + 20 = -x + 4

y = -x - 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

y - (-20) = 1 (x - 4)

y + 20 = x - 4

y = x - 24

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1 year ago

GaryK [48]

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see the explanation

Step-by-step explanation:

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2 years ago

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Nuetrik [128]

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<h3>What is the linear system?</h3>

A linear system is one in which the parameter in the equation has a degree of one. It might have one, two, or even more variables.

A coordinate plane. The x-axis and y-axis are each scaled by one.

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More about the linear system link is given below.

brainly.com/question/20379472

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1 year ago

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Volgvan

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Let X the random variable that represent the scores, and for this case we know the distribution for X is given by:

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