Maths problem

If you use a form of the quadratic that looks like:

y = a(x-b)(x-c)

Then b and c are the x intercepts. You can put any value in for the constant a you want and get different curves. This will work for other polynomials as well.

Let the two equations be

y=a1x ^2 + b1x +c1 =0
y=a2x ^2 + b2x +c2 =0

for intersection solve by quadratic formula
(a1-a2) x ^2 + (b1-b2)x +(c1-c2) =0

Any polynomial can written as using cofficients a,b,c,d etc on the powers of x,

e.g. y=ax^n + bx^(n-1) + cx^(n-2) +......+ zx^(n-25)+.....etc

hey ho

I have a similer problem.

my two x-int the same:

x1= -2
x2= 2

They are shared by to graphs, one is a quadratic and the other is a polynomial. So far I get them to intercect on the x-int's through trial and error, but I was wondering if there is a algerbraeic formula to calculate it.

Also, is there part of the function that determines the x-intercept? Like the 'c' value = y intercept.

Yes, just expand my answer from earlier.

Your quadratic is

y = a (x-2) (x+2)

This has roots at 2 and -2 for any value of a you put in. For your polynomial, use...

y = b (x-2)(x+2)(x-c)(x-d)...

Where c and d are other roots and b is any constant you want. If you want higher order polynomials, just add more terms.

Ah create equations !

Cheers mate,

one other thing -> I understand the quadratic 'a(x+2)(x-2)'

but in the polynomial -> what do 'c' and 'd' stand for? I'm assuming that 'a' and 'b' stand for the x-int.

btw thanks for your help

a and b are constants. They can be anything you choose and not change the x intercepts. c and d stand for other intercepts. If you use c and d, you get a fourth order polynominal. If you use c, d, and e, you would get a fifth order, etc. You could also put any polymonial you want in place of c and d and still get the desired results. For example:

The quadratic equation 3(x-2)(x+2) will share x intercepts with -2(x-2)(x+2)(x-5).

I just used one additional intercept (5) to produce a third order polynomial. I used 3 and -2 as the constants in front of the equation.

The quadratic equation 5(X-2)(X+2) will share x intercepts with -(x+2)(x-2)(x^2 + 1)

Here I just tacked a polynomial on the end (x^2+1) and used 5 and -1 as the constants a and b.

Ok i got that now, thankyou very much. Your method worked like a charm. by the way, to change -(x+2)(X-2)(x^2+1) into an 'ax+bx+c' form, do you just expand the brackets? When I use the expanded form i get a different graph altogether.

The bracketed format works perfectly, except I can't go to the next step without finding how to change it that ax+bx+c type format.

Yes, you should just be able to expand it by multiplying it out. If you are getting a different graph, there is probably a simple error in the expansion somewhere. In the example, -(x+2)(X-2)(x^2+1) expands out to -x^4 + 3x^2 + 4.