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write an equation of a polynomial with x intercepts of 0, 10, -15 and passes through the point (2, -136)

User Freek
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1 Answer

4 votes

Answer:

f(x) = 0.5x^3 + 2.5x^2 - 75x

Explanation:

To write the equation of a polynomial with given x-intercepts and passing through a point, we can use the factored form of the polynomial:

f(x) = a(x - r1)(x - r2)(x - r3) ...

where r1, r2, r3, ... are the x-intercepts, and a is a constant that scales the polynomial vertically. Since the x-intercepts are 0, 10, and -15, we can write:

f(x) = a(x - 0)(x - 10)(x + 15)

Expanding this expression gives:

f(x) = a(x^3 + 5x^2 - 150x)

To find the value of a, we can use the fact that the polynomial passes through the point (2, -136):

-136 = a(2^3 + 5(2^2) - 150(2))

Simplifying this equation gives:

-136 = -272a

Dividing both sides by -880 gives:

a = 0.5

Therefore, the equation of the polynomial is:

f(x) = 0.5x^3 + 2.5x^2 - 75x

User MultiRRomero
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