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2 years ago

PLEASE HELP

Mathematics

1 answer:

SVETLANKA909090 [29]2 years ago

6 0

Answer:

Hill 1: F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)

Hill 2: F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)

Hill 3: F(x) = 4(x - 2)(x + 5)

Step-by-step explanation:

Hill 1

You must go up and down to make a peak, so your function must cross the x-axis six times. You need six zeros.

Also, the end behaviour must have F(x) ⟶ -∞ as x ⟶ -∞ and F(x) ⟶ -∞ as x⟶ ∞. You need a negative sign in front of the binomials.

One possibility is

F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)

Hill 2

Multiplying the polynomial by -½ makes the slopes shallower. You must multiply by -2 to make them steeper. Of course, flipping the hills converts them into valleys.

Adding 3 to a function shifts it up three units. To shift it three units to the right, you must subtract 3 from each value of x.

The transformed function should be

F(x) = -2(x +1)(x)(x -2)(x -3)(x - 6)(x - 7)

Hill 3

To make a shallow parabola, you must divide it by a number. The factor should be ¼, not 4.

The zeroes of your picture run from -4 to +7.

One of the zeros of your parabola is +5 (2 less than 7).

Rather than put the other zero at ½, I would put it at (2 more than -4) to make the parabola cover the picture more evenly.

The function could be

F(x) = ¼(x - 2)(x + 5).

In the image below, Hill 1 is red, Hill 2 is blue, and Hill 3 is the shallow black parabola.

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