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denpristay [2]
3 years ago
9

The graph of which function will have a maximum and a y-intercept of 4? f(x) = 4x2 + 6x – 1 f(x) = –4x2 + 8x + 5 f(x) = –x2 + 2x

+ 4 f(x) = x2 + 4x – 4
Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
7 0

Answer:

It is f(x) = -x^2 + 2x + 4.

Step-by-step explanation:

As it has a maximum the coefficents of x^2 will be negative.

f(x) = -x^2 + 2x + 4

The y-intercept occurs when x = 0 so

y-intercept = 0 + 0 + 4 = 4.

-

Yakvenalex [24]3 years ago
4 0

ANSWER

f(x) =  -  {x}^{2}  + 2x + 4

EXPLANATION

The graph of the function that will have a maximum and a y-intercept of 4 is

f(x) =  -  {x}^{2}  + 2x + 4

To check for the y-intercept, we put x=0 into the function.

f(0) =  -  {0}^{2}  + 2(0)+ 4

This implies that,

f(0) = 4

We can see that the coefficient of the quadratic term is -1.

Since the coefficient of the quadratic term is less than zero, the graph will have a maximum point.

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a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}

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\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then a_n will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When n=2, you have

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Assume a_k\ge a_{k-1}, i.e. that a_k=\sqrt{2a_{k-1}}\ge a_{k-1}. Then for n=k+1, you have

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Clearly, a_1=2^{1/2}. Let's assume this is the case for n=k, i.e. that a_k. Now for n=k+1, we have

a_{k+1}=\sqrt{2a_k}

and so by induction, it follows that a_n for all n\ge1.

Therefore the second sequence must also converge (to 2).
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