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Identify the quadratic function that is in standard form has zeros 4 and −7.

A. f(x)=x²+3x+28
B. f(x)=x²+11x-28
C. f(x)=x²+3x-28
D. f(x)=x²-11x+28

User HeLomaN
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2 Answers

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Step-by-step explanation:

The standard form of a quadratic function is given by \(f(x) = ax^2 + bx + c\). The zeros of the function are the values of \(x\) for which \(f(x) = 0\).

If the zeros are \(x = 4\) and \(x = -7\), the corresponding factors would be \((x - 4)\) and \((x + 7)\). Therefore, the quadratic function in standard form would be:

\[f(x) = a(x - 4)(x + 7)\]

Now, expand and simplify to determine the values of \(a\):

\[f(x) = a(x^2 + 7x - 4x - 28)\]

Combine like terms:

\[f(x) = a(x^2 + 3x - 28)\]

Now, compare this with the given options:

A. \(f(x) = x^2 + 3x + 28\) - Incorrect

B. \(f(x) = x^2 + 11x - 28\) - Incorrect

C. \(f(x) = x^2 + 3x - 28\) - Correct

D. \(f(x) = x^2 - 11x + 28\) - Incorrect

Therefore, the quadratic function in standard form with zeros 4 and -7 is \(f(x) = x^2 + 3x - 28\), so the correct answer is option C.

User Lifeless
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4 votes

Final answer:

The quadratic function in standard form with zeros 4 and -7 is f(x) = (x - 4)(x + 7), which expands to f(x) = x^2 + 3x - 28.

Step-by-step explanation:

The quadratic function in standard form with zeros 4 and -7 is f(x) = (x - 4)(x + 7).

In standard form, a quadratic function is written as f(x) = ax^2 + bx + c. Since the zeros are given, we can write the function as f(x) = (x - 4)(x + 7), which when expanded gives f(x) = x^2 + 3x - 28.

Therefore, the correct answer is option C, f(x) = x^2 + 3x - 28.

User Basant Singh
by
8.4k points

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