Question

Input

Answer 1

If we graph the given points then they appear to be in the shape of cubic function so we can use standard formula of cubic function which is

$y=ax^3+bx^2+cx+d$

Plug the given points to get four equations

like first point (3,6) so plug x=3 and y=6 into above equation, we get:

$6=a*3^3+b*3^2+c*3+d$

or

6=27a+9b+3c+d

Same way by plugging other points form table, we get total four equations as follows:

6=27a+9b+3c+d, 11=125a+25b+5c+d, 11=64a+16b+4c+d, 21=343a+49b+7c+d

Now we just need to solve these equations for a, b, c and d.

Solving them to get values of a, b, c and d, we get:

$a=\frac{25}{24}, b=-15, c=\frac{1715}{24}, d=-\frac{203}{2}$

Now plug them into

$y=ax^3+bx^2+cx+d$

So the required equation will be:

$y=\frac{25}{24}x^3-15x^2+\frac{1715}{24}x-\frac{203}{2}$

Hence required equation in function notation can be written as

$f(x)=\frac{25}{24}x^3-15x^2+\frac{1715}{24}x-\frac{203}{2}$

Now to prove that above function is correct, we just graph the given points from table and the obtained function.

After graphing the given points from table and the obtained function, We see that points lie on the graph of $y=\frac{25}{24}x^3-15x^2+\frac{1715}{24}x-\frac{203}{2}$

Which proves that our equation is correct.