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

Answer:

$y =-\frac{1}{2}(x +2)(x - 3)$

Step-by-step 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

$k=\frac{2}{-4}$

$k=-\frac{1}{2}$

So, the equation is:

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

$y =-\frac{1}{2}(x +2)(x - 3)$