In the given triangles, PR ≅ AC and ∠P ≅ ∠A.
1 answer:
You might be interested in
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.
