Which function is a quadratic function? y – 3x2 = 3(x2 + 5) + 1 y2 – 7x = 2(x2 + 6) + 7 y – 2x2 = 6(x3 + 5) – 4 y – 5x = 4(x + 5
1 answer:
Answer:
Option A.
Step-by-step explanation:
The general form of a quadratic function is
It means, power of y should be 1 and highest power of x should be 2.
In option A,
It is a quadratic function.
In option B,
Here, power of y is 2. So, it is not be a quadratic function.
In option C,
Here, highest power of x is 3. So, it is not be a quadratic function.
In option D,
Here, highest power of x is 1. So, it is not be a quadratic function.
Therefore, the correct option is A.
You might be interested in
I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take 
, so that 
and we're left with the ODE linear in 
:
Now suppose
has a power series expansion
Then the ODE can be written as
![\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%5Cge2%7D%5Cbigg%5Bn%28n-1%29a_n-%28n-1%29a_%7Bn-1%7D%5Cbigg%5Dx%5E%7Bn-2%7D%3D0)
All the coefficients of the series vanish, and setting
in the power series forms for and 
tell us that 
and 
, so we get the recurrence
We can solve explicitly for
quite easily:
and so on. Continuing in this way we end up with
so that the solution to the ODE is
We also require the solution to satisfy, which we can do easily by adding and subtracting a constant as needed:
Now suppose
Then the ODE can be written as
All the coefficients of the series vanish, and setting
We can solve explicitly for
and so on. Continuing in this way we end up with
so that the solution to the ODE is
We also require the solution to satisfy