Math JAVA challenging question question. Please see the pictures and there is th

Question

Math JAVA challenging question question. Please see the pictures and there is the question and the library. The questions and the references to the libraries are included!! JUST ANSWER THE QUESTION WORTH 5 points I ALREADY ATTEMPTED IT!!

1limport 4 java . util . ArrayList; 5 public class LibrariesChallengePS 6 7 8 7 5 points 9 static Polynomial Q1 (Polynomial input, double x)f 10 11 double x1 = input . evaluate(x); 12 13 14 Polynomi alp= new Polynomial(list); 15 16 17 18 // return any Polynomial that intersects the input Polynomial at x ArrayListDouble> list new ArrayList(); = return null;

Explanation / Answer

Suppose a polynomial is given to you as:

x^2 -3x - 4 = 0;

If you try finding polynomial which intersects it at x = 6, Then both the polynomials should have the same value at x = 6..

So value of x^2 -3x - 4 at x=6 is: 36 - 18 - 4 = 14..

So we are sure that at x = 6, the resultant polynomial should give us value 14..

Now There can be a lot of polynomials like line, circle, parabola etc. which may give us value of 14 at x = 6.. x=6 and value of 14 is nothing but just a point (6,14) and there can be inifinite shapes / lines passing through a point in the space..

A simple line can be a line passing parallel to x axis which always returns value 14 on y axis.. Its equation will be y = 14.. you notice that there is no power of x in the equation, hence it will not change its y value at any value of x.

Similarly, there can be line with gradient(not parallel to x axis) which are passing through point (6,14)..

a simple example can be y = x+8, This line will give you y = 14 at x = 6..

There can be many more examples..

For this question, lets try to answer in form of polynomial as x + c, where c is the constant term..
As given in above example, if we need to find intersection at p, and the value of Poynomial comes as v at x=p, then we can find the equation as x + (v - p)
in the last example, p was 6, and v came to 14, hence the derived equation was x + (14 - 6) = x - 8(Answer)

now comes the progamming part:

static Polynomial Q1(Polynomial input, double p) {

double v = input.evaluate(p);

ArrayList<Double> list = new ArrayList<Double>();

// add constant term

list.add(v - p);

// add the term containing x power..

// notice that multiple of x will always be 1 in this case

list.add(1);

return new Polynomial(list);

}

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---