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Given the following zeros and a point on the graph, write the equation in standard form.

(-2, 0), (-4, 0), and (0, 16)
a. y = -2x^2 + 4x + 16
b. y = 2x^2 + 4x - 16
c. y = -2x^2 - 4x + 16
d. y = 2x^2 - 4x - 16

User Yussan
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1 Answer

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Final answer:

Given the zeros (-2, 0) and (-4, 0), the factors of the polynomial are (x + 2) and (x + 4), which multiply to x^2 + 6x + 8. Using the point (0, 16), we find that the leading coefficient is 2, resulting in the equation y = 2x^2 + 6x + 16. The provided options do not match this equation, with option (b) being the closest if we ignore a typo in the constant term.

Step-by-step explanation:

To write the equation of a polynomial given its zeros and a point on its graph, we start by using the zeros to form the factors of the polynomial. Since the zeros are (-2, 0) and (-4, 0), the factors of the polynomial are (x + 2) and (x + 4). Multiplying these factors together gives us the quadratic in the form (x + 2)(x + 4), which simplifies to x^2 + 6x + 8.

Next, because we know another point on the graph is (0, 16), we can use this to find the leading coefficient. Substituting x = 0 into the equation y = ax^2 + bx + c and setting y to 16, we obtain c = 16, as the leading coefficient must be a number that multiplies the x^2 term to satisfy the point (0, 16). However, since the constant term resulting from the multiplication of the factors is 8, and we need c to be 16, the leading coefficient must be 2 to double the constant term (2*8 = 16).

Therefore, the equation in standard form with a leading coefficient of 2 is y = 2x^2 + 6x + 16. Comparing this with the given options, we can correct a typo and see that none of the options are correct, and the correct equation should be y = 2x^2 + 6x + 16, not an option given. However, if the error in the constant term is a simple typo, the correct choice closest to our derived equation would be option (b) y = 2x^2 + 4x - 16, if we ignore the error in the constant term.

User Tomek Miszczyk
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