Louis bought a laptop for $720. It was marked $90 off because it was an open product. He also got a 15% discount, which was take
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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.
The intercepts of the third degree polynomial corresponds to the zeros of the equation
y = d*(x-a)*(x-b)(x-c)
Where a, b and c are the roots of the polynomial and d an adjustment coefficient.
y = d*(x+2)*(x)*(x-3)
Lets assume d = 1, and we get
y = (x+2)*(x)*(x-3) = x^3 - x^2 - 6x
We graph the equation in the attached file.
y = d*(x-a)*(x-b)(x-c)
Where a, b and c are the roots of the polynomial and d an adjustment coefficient.
y = d*(x+2)*(x)*(x-3)
Lets assume d = 1, and we get
y = (x+2)*(x)*(x-3) = x^3 - x^2 - 6x
We graph the equation in the attached file.