# 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

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)

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1 year 2021-09-05T08:55:18+00:00 1 Answers 193 views 0

1. Given:

The image of a lens crosses the x-axis at –2 and 3.

The point (–1, 2) is also on the parabola.

To find:

The equation that can be used to model the image of the lens.

Solution:

If the graph of polynomial intersect the x-axis at c, then (x-c) is a factor of the polynomial.

It is given that the image of a lens crosses the x-axis at –2 and 3. It means (x+2) and (x-3) are factors of the function.

So, the equation of the parabola is:

[tex]y=k(x+2)(x-3)[/tex]          …(i)

Where, k is a constant.

It is given that the point (–1, 2) is also on the parabola. It means the equation of the parabola must be satisfy by the point (-1,2).

Putting [tex]x=-1, y=2[/tex] in (i), we get

[tex]2=k(-1+2)(-1-3)[/tex]

[tex]2=k(1)(-4)[/tex]

[tex]2=-4k[/tex]

Divide both sides by -4.

[tex]\dfrac{2}{-4}=k[/tex]

[tex]-\dfrac{1}{2}=k[/tex]

Putting [tex]k=-\dfrac{1}{2}[/tex] in (i), we get

[tex]y=-\dfrac{1}{2}(x+2)(x-3)[/tex]

Therefore, the required equation of the parabola is [tex]y=-\dfrac{1}{2}(x+2)(x-3)[/tex].

Note: All options are incorrect.