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3 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]3 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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<h3>Answer: Point E is located at (2, 3)</h3>

====================================

Explanation:

The sequence PQRST has T as the last letter.

The sequence ABCDE has E as the last letter

The two letters T and E pair up or they correspond. So T will rotate to land on point E. Or T will rotate to point E's location.

Point T is at location (-3,2). So x = -3 and y = 2.

-------

Use the rule

(x,y) ---> (y, -x)

which is the rule we use when we rotate any point 90 degrees clockwise around the origin. We swap the x and y coordinates, then change the sign of the second coordinate (after the swap happens).

This means,

(-3, 2) ----> (2, -(-3)) = (2, 3)

or simply

(-3, 2) ---> (2, 3)

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