The given are Two x-intercepts y-intercept math (0,−4) Maximum at (2,4) How to create quadratic equation given y intercept, and maximum and B=8?

Lyla Carson 2022-09-24 Answered
The given are
Two x-intercepts
y-intercept math ( 0 , 4 )
Maximum at ( 2 , 4 )
How to create quadratic equation given y intercept, and maximum and B = 8?
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Answers (2)

embraci4i
Answered 2022-09-25 Author has 10 answers
You can start by writing out the general form of a quadratic polynomial: P ( x ) = a x 2 + b x + c. Saying that the y-intercept is ( 0 , 4 ) is the same as saying that P ( 0 ) = 4, which gives us c = 4. Now the derivative of the polynomial will be P ( x ) = 2 a x + b. Since the derivative of a differentiable function at a maximum is 0, you must have P ( 2 ) = 0 or 4 a = b. Since the point ( 2 , 4 ) is assumed to be in the graph of the function, we know that P ( 2 ) = 4 or 4 a + 2 b 4 = 4, which means that 4 a = 8 2 b. Thus we have 4 a = b and   4 a = 8 2 b
Which gives b = 8 and a = 2, so we get P ( x ) = 2 x 2 + 8 x 4
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deiluefniwf
Answered 2022-09-26 Author has 4 answers
Another way to do it, would be to use the fact that every quadratic equation is of the form P ( x ) = α ( x β ) 2 + γ for some α , β , γ. When the equation is written in this form it becomes clear that α must be negative (otherwise the function won't have a maximum), and then the maximum is attained when x β = 0. Since the maximum is at x = 2, we have β = 2. Also, the value of P at the maximum will be 4 = P ( 2 ) = α ( 0 ) 2 + γ = γ, so γ = 4. Lastly, we know that the y-intercept is ( 0 , 4 ), so that 4 = P ( 0 ) = α ( 0 β ) 2 + γ = 4 α + 4, which gives α = 2. Now we can expand the original expression and we get P ( x ) = α ( x β ) 2 + γ = 2 ( x 2 ) 2 + 4 = 2 x 2 + 8 x 4, so we get the same result as before.
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