Stationary points. Help ASAP please. Thanks

You can use the power rule for derivatives for each problem (though you could also use the product rule for the fourth curve).

y = x ² + 6x - 1   ==>   dy/dx = 2x + 6

y = x ² - 5x + 1   ==>   dy/dx = 2x - 5

y = 2 - 4x - x ²   ==>   dy/dx = -4 - 2x

y = (1 + x) (7 - x) = 7 + 6x - x ²   ==>   dy/dx = 6 - 2x

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or, using the product rule,

dy/dx = (1 + x) (-1) + 1 (7 - x) = -1 - x + 7 - x = 6 - 2x

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Now, stationary points occur where the derivative is zero. We have

2x + 6 = 0   ==>   x = -3

2x - 5 = 0   ==>   x = 5/2

-4 - 2x = 0   ==>   x = -2

6 - 2x = 0   ==>   x = 3

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option