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)

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:

         ...(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 in (i), we get

Divide both sides by -4.

Putting in (i), we get

Therefore, the required equation of the parabola is .

Note: All options are incorrect.

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

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step-by-step explanation: