The second-degree polynomial function f(x) that has a lead coefficient of 4 and roots 5 and 2 is f(x) = 4x² -28x + 40. The correct option is the last option f(x)=4x²-28x+40
<h3>Quadratic equation</h3>
From the question, we are to determine which second-degree polynomial function f(x) has a lead coefficient of 4 and roots 5 and 2
To determine this, we will use the given roots to determine the equation
The roots are 5 and 2
Thus
x = 5 and x = 2
x - 5 = 0 and x - 2 = 0
Therefore,
(x -5)(x -2) = 0
Distributing
x(x -2) -5(x -2) = 0
x² -2x -5x + 10 = 0
x² -7x + 10 = 0
Now, multiplying through by 4, we get
4x² -28x + 40 = 0
Thus, the function becomes f(x) = 4x² -28x + 40
Hence, the second-degree polynomial function f(x) that has a lead coefficient of 4 and roots 5 and 2 is f(x) = 4x² -28x + 40. The correct option is the last option f(x)=4x²-28x+40
Learn more on Quadratic Equation here: brainly.com/question/19257086
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