A polynomial has zeros of 7, 6. What is the minimum degree of the polynomial?
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Part A. The technique on how to find the equation that only applies to point D and E, is to create a line or curve that only includes two of these points. In this case, I created a random parabola that isolates points C and F from the rest of the points. First, we have to find the equation of the parabola through its general forms:
(x - h)² = +/-4a(y-k) or (y - k)² = +/-4a(x - h)
For parabolas drawn like that in the picture, the general form is (x - h)² = +4a(y-k), where the vertex is (h,k) and a is the distance from the vertex to the focus. From the picture, the vertex is at (0,3). Then, we use point D(-2,4) to determine a:
(-2 - 0)² = +4a(4 - 3)
4a = 4
So, the equation of the parabola is:
x² = 4(y - 3)
x² = 4y - 12
Part B. Point D was already verified above. Now for point E(2,4)
x² = 4y - 12
2² ? 4(4) - 12
4 ? 4
4 = 4
Part C. For y < 7x − 4, ignore the equality symbol first and graph the line. Assign random values of x, then you get corresponding values of y. Plot them as shown in the second picture. The line is shown in red. Next, test the equation by choosing a random point. Let's choose the purple point at (4,3).
3 ? 7(4) − 4
3 ? 24
3 < 24
Thus, it applies to the purple point, and all the other areas to that area. The shaded region are all solutions of the inequality. So, Erica is only interested in points E, C and F.
(x - h)² = +/-4a(y-k) or (y - k)² = +/-4a(x - h)
For parabolas drawn like that in the picture, the general form is (x - h)² = +4a(y-k), where the vertex is (h,k) and a is the distance from the vertex to the focus. From the picture, the vertex is at (0,3). Then, we use point D(-2,4) to determine a:
(-2 - 0)² = +4a(4 - 3)
4a = 4
So, the equation of the parabola is:
x² = 4(y - 3)
x² = 4y - 12
Part B. Point D was already verified above. Now for point E(2,4)
x² = 4y - 12
2² ? 4(4) - 12
4 ? 4
4 = 4
Part C. For y < 7x − 4, ignore the equality symbol first and graph the line. Assign random values of x, then you get corresponding values of y. Plot them as shown in the second picture. The line is shown in red. Next, test the equation by choosing a random point. Let's choose the purple point at (4,3).
3 ? 7(4) − 4
3 ? 24
3 < 24
Thus, it applies to the purple point, and all the other areas to that area. The shaded region are all solutions of the inequality. So, Erica is only interested in points E, C and F.