Write an equation of the parabola in intercept form that passes through (0, −18) with x-intercepts of 9 and 1.

Answer 1

Parabola equation is in vertex form it's y-32=-2(x-5)²

or in y-intercept form y=-2x²+20x-18. The constant term, -18, is the y-intercept

the x-intercepts are the values of x that make x-5 = absolute value of 4, or 9-5 and 1-5.

The midpoint between the x intercepts is the line of symmetry. x=5 is that line of symmetry which has the vertex. 5 is the midpoint between 9 and 1. The parabola opens below, with general formula y-k=a(x-h) with vertex (h,k) or (5,k) as h=5,

k > 0 and a<0

You have 3 points on the parabola, (0,-18), (9,0) and (1,0). the vertex is also a point on the parabola at (5,k) where k is the maximum point on the parabola, greater than zero. Take the first derivative of the parabola equation and set it = zero. y'=2ax-10a=0 cancel the a's to get 2x-10=0 x=5

So far we just have y-k=a(x-5)². plug one x-intercept, one

point into the general formula: (1,0) to get 0-k=a(1-5)² or -k=16a

or 16a=-k

Plug another point into the formula, (0,-18) to get -18-k=a(0-5)² or 25a=-18-k

subtract from that last equation, 16a=-k to get 25a-16a=-18-k+k or 9a=-18 or a=-2, then k=-16a=-16(-2)=32

the formula for the parabola is now (y-32)=-2(x-5)² or y-32=-2x²+20x-50

the vertex is (5,32). Plug it into the parabola equation, or any of the other 3 known points, and the equation works. 32-32=-2(5)²+20(5)-50=-50+100-50=0 Try each x-intercept: 0-32=-2(9)²+20(9)-50=-32. Plug in the y-intercept, (0,-18) -18-32=0+0-50=-50

The parabola opens downward with vertex, the maximum point at (5,32). (0,-18) is the y-intercept, and the two x-intercepts are given as 9 and 1 or (1,0) and (9,0)