Question

The image of a parabolic lens is projected onto a graph. The image crosses the x-axis at –2 and 3. The point (–1, 2) is also on the parabola. Which equation can be used to model the image of the lens?

y = (x – 2)(x + 3)
y = (x – 2)(x + 3)
y = (x + 2)(x – 3)
y = (x + 2)(x – 3)

Answer 1

C

Explanation:

Answer 2

`y =-(1)/(2)(x +2)(x - 3)`

Explanation:

Given

`x_1 = -2`

`x_2 = 3`

`(x,y) = (-1,2)` --- a point on the parabola

Required

The equation

First, calculate the equation from the zeros

`y =k(x - x_1)(x - x_2)`

Substitute
`x_1 = -2` and
`x_2 = 3`

`y =k(x - -2)(x - 3)`

`y =k(x +2)(x - 3)`

To solve for k, we substitute
`(x,y) = (-1,2)`

`2 = k(-1+2)(-1-3)`

`2 = k(1)(-4)`

`2 = -4k`

Divide by -4

`k=(2)/(-4)`

`k=-(1)/(2)`

So, the equation is:

`y =k(x +2)(x - 3)`

`y =-(1)/(2)(x +2)(x - 3)`