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What is the location of the vertex in the following function,

f(x) = 32(x + 3y^2 - 5) ?

a) (-3, -5)
b) (3, 5)
c) (-3, 5)
d) (3, -5)

User JGleason
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1 Answer

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Final answer:

The vertex of a function in vertex form is at the point (h, k). Assuming a typo in the question and correcting it to a quadratic function of x only, the vertex would be at (-3, -5). However, without a corrected function, we cannot definitively determine the vertex.

Step-by-step explanation:

The vertex of a function in vertex form, f(x) = a(x - h)^2 + k, is located at the point (h, k). The function provided, f(x) = 32(x + 3y^2 - 5), is not properly written to represent a function in terms of x only, as it includes a y^2 term. Assuming there is a typo and the function should be quadratic in form of x, such as f(x) = a(x - h)^2 + k, and the y term is actually meant to be part of either h or k, we can identify the vertex. If we treat the y term as a typo and focus on the x term only, the function would resemble f(x) = 32(x + 3)^2 - 5 after correction, and therefore the vertex would be at (-3, -5). Without more context or correction to the function, we cannot definitively determine the vertex location.

User Sebastian Gomez
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