Philip P. answered • 04/15/15

Affordable, Experienced, and Patient Algebra Tutor

A degree 2 polynomial is a quadratic equation whose graph is a parabola. If the range is (-∞,9), then we have an inverted parabola with the vertex at the top. The y-coordinate of the vertex is 9. Use the vertex form of the quadratic equation:

y = a(x-h)

^{2}+ kwhere a is a constant and (h,k) is the location of the vertex. So k = 9 and so far we have:

y = a(x-h)

^{2}+ 9We need to find a and h. h is the x-coordinate of the vertex. It lies exactly half way between the x-intercepts, which are -1 and 5, a distance of 6 apart. Half of 6 is 3, so the vertex's x-coordinate, h = -1 + 3 = 2. Now our equation is:

y = a(x-2)

^{2}+ 9To find a, let's plug in one of the x-intercept locations, say (-1,0):

0 = a(-1-2)

^{2}+ 90 = a(-3)

^{2}+ 90 = 9a + 9

-9 = 9a

-1 = a

So the final equation in

**vertex form**is:**y = -(x-2)**

^{2}+ 9In

**standard form**, it's:**y = -x**

^{2}+ 4x + 5In

**factored form**, it's**y = -(x-5)(x+1)**