Equation of a polynomial function

Question

Find the equation of a polynomial function of degree $4$ whose graph is shown in the figure. I found $c=0$ and $b=-4a$ but I cant go further

Answer 1

Polynomial of degree $ 4$:

Zeroes: $ -2,-1, 2$.

Double zero at $x=2$ (Why?)

Ansatz:

$y=a(x-2)^2(x+2)(x+1);$

At $x=0, y=4;$

Determine $a.$

Recall:

A polynomial, real coefficients, of degree $4$ has $4$ roots.There are $3$ real roots, why is the $4$th root real? And where is it in the drawing?

Answer 2

We have $f(x)=ax^4+bx^3+cx^2+dx+e$.

Furthermore:

$f(0)=4, f(2)=0, f(-2)=0, f'(-1)=0$ and $f'(2)=0.$

This gives five equations for the unknown $a,b,c,d, e.$

Answer 3

Hint:

1. Let $f(x)=\displaystyle\sum_{k=0}^4 a_kx^k=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0$.
2. From graph $-2, -1, 2$ are zeroes, this means $(x+2)(x+1)(x-2)$ is factor of $f(x)$.
3. Also, $f$ intercepts the $y$-axis at $y=4$, which implies $f(0)=4$.
4. Differentiate $f(x)$ and look for points where the curve has a horizontal tangent or a slope of $0$. Clearly, it happens at $x=2$, which implies $f'(x)=4a_4x^3+3a_3x^2+2a_2x+a_1$ has a zero at $x=2$.

This gives you a couple of linear equations to work with solving which you get the values of $\{a_i \}_{i=0}^4$, plug them back into $f(x)$ and you are done.

Answer 4

The function of a fourth degree polynomial is $f(x)=ax^4+bx^3+cx^2+dx+e$ or $f(x)=a(x-r_1)(x-r_2)(x-r_3)(x-r_4)$ in factored form, with $r_i$ being the solutions of the polynomial and $a$ the general coefficient.

Here, the polynomial is $f(x)=a(x+2)(x+1)(x-2)(x-2)$ (double solution at $x=2$).

Plugging $x=0$ makes our equation $4=a(2)(1)(-2)(-2)$ ($4$ is from the point $(0,4)$, so $a=1/2$

Finally our polynomial is $$f(x)=\frac{1}{2}(x+2)(x+1)(x-2)(x-2)$$

Expand this and we are done.