Question

Find the equation defining graphs (a), (b) and (c).
You can express it in any of the 3 ways: general form, vertex form or factorised form (if possible).
Show all the working out.
Upload your work in the managebac dropbox.
Note: exercise (d) is an extension. A bonus will be given if done correctly.
Hint: A system of 3 simultaneous equations must be solved with the GDC.

Answer 1

Respuesta:

Explicación paso a paso:

Question (a):

The graph of this function is a convex parabola, and it got one y-intercept in the point (0,4) and two x-intercept in the points (1,0) and (5,0). Then let put these points in the general formula:

Remember that the form of this function is: f(x)=ax²+bx+c, with a∈R, b∈R, c∈R, so let use the points in that formula:

f(x)=ax²+bx+c

*Point (0,4):

4=a(0)²+b(0)+c

c=4

*Point (1,0):

0=a(1)²+b(1)+c

0=a+b+c

*and point (5,0)

0=a(5)²+b(5)+c

0=25a+5b+c

Look that the first point (0,4) say that c=4, then let c=4 in the others two points:

0=a+b+4

0=25a+5b+4

You see a linear equations system with two variables and two equations. Let resolve by the substitution method.

a+b=-4

25a+5b=-4

b=-a-4

25a+5(-a-4)=-4

25a-5a-20=-4

20a-20=-4

20a=16

a=16/20=4/5

And if a=4/5, c=4, then b=-s-4=-4/5-4=-24/5

So the function is f(x)= (4/5)x²-(24/5)x+4

I hope you can do the others exercises. :)