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What is the y-value of the vertex of the function f(x)=-(x-3)(x+11)?

a. -8
b. -4
c. 33
d. 49

User Cjbarth
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1 Answer

3 votes

Final answer:

The y-coordinate of the vertex for the function f(x) = -(x-3)(x+11) is found to be -15 after expanding and substituting the x-coordinate of the vertex into the function. However, this value does not match any of the provided options.

Step-by-step explanation:

The y-value of the vertex of the quadratic function f(x) = -(x-3)(x+11) can be found by first expressing the function in standard form, which is f(x) = ax^2 + bx + c, where the vertex's x-coordinate is given by -b/(2a). To find the y-coordinate of the vertex, we substitute this value into the function. Expanding f(x) gives:

  • f(x) = -[(x^2 + 11x) - (3x + 33)]
  • f(x) = -[x^2 + 8x - 33]

In this form, a = -1, b = -8, and the vertex x-coordinate is x = -(-8)/(2 × (-1)) = 4. Plugging this back into the function:

  • f(4) = -[(4)^2 + 8(4) - 33]
  • f(4) = -[16 + 32 - 33]
  • f(4) = -[48 - 33]
  • f(4) = -15

Therefore, the y-coordinate of the vertex is -15, which is not listed in the provided options, implying there might be a misunderstanding in the question or the options given.

User Geoff Clayton
by
7.8k points

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