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Find the quadratic function f(x) that goes through (0, 0) and has a local at (0,).

a) f(x) = x²
b) f(x) = -x²
c) f(x) = x² + 1
d) f(x) = -x² - 1

User Rethinavel
by
8.4k points

1 Answer

6 votes

Final answer:

The quadratic function f(x) is either x^2 if the local extremum at the origin is a minimum or -x^2 if it is a maximum. Since the question lacks the detail about the nature of the extremum, either of these functions could be correct.

Step-by-step explanation:

The student is asking for the quadratic function f(x) that passes through the point (0, 0) and has a local extremum at (0, y). Since the point (0, 0) is on the graph of the function, the constant term c in the standard form of the quadratic equation ax^2 + bx + c = 0 must be zero.

Given the options, only functions a) f(x) = x^2 and b) f(x) = -x^2 go through the origin (0, 0). However, as the question specifies that there is a local extremum at (0, y), which must be the vertex of the parabola, we must determine whether it is a maximum or a minimum.

To find the quadratic function that goes through (0,0) and has a local minimum at (0, ?), we can use the form f(x) = ax^2 + bx + c. Since the vertex of the parabola is at (0, ?), the x-coordinate of the vertex is 0. Plugging this into the equation, we get:

f(0) = a(0)^2 + b(0) + c = c

Since the y-coordinate of (0, ?) is 0, we have c = 0. Therefore, the quadratic function is f(x) = ax^2 + bx.

This detail in the question is incomplete; therefore, based solely on the given information, either f(x) = x^2 if the extremum is a minimum or f(x) = -x^2 if it is a maximum could be the correct answer.

User BSKANIA
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