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
[Math] Equation of a polynomial function
functionspolynomials
Related Solutions
The general equation for a straight line in Cartesian co-ordinates is $y=mx+c$, where $m$ is the gradient of the straight line and $c$ is the y-intercept of the line.
In the case of your line $f(x)$, you have $m=3$, $c=5$. So you have a straight line which passes through the point $(0,5)$ and has gradient $3$, this means you can plot a graph like that shown below (note we can calculate the x-intercept as $(-\frac{5}{3},0)$ by setting $f(x)=0$):
Hope this helps!
note:corrected the x-intercept to (-5/3,0)
You can do this most easily with matrices, and a quick Google search shows that matrix multiplication can be done in Excel, but it can be a bit of a pain, especially with so many points.
Let $A$ be the matrix created from the system of equations that result from plugging each point into $f(x)$ and $y$ be the matrix containing the values of $f(x)$. So given a set of points $\{(x_1, y_1),(x_2, y_2),...,(x_n, y_n)\}$, if you want a polynomial fit of degree $N$ (with $N>n$), you'd have the following matrices:
$A = \begin{bmatrix} x_1^{N} & x_1^{N-1} & x_1^{N-2} & ... & 1 \\ x_2^{N} & x_2^{N-1} & x_2^{N-2} & ... & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_n^{N} & x_n^{N-1} & x_n^{N-2} & ... & 1 \end{bmatrix}$
$Y = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}$
Now take set the matrix $\beta$ as
$$\beta = (A^T A)^{-1} A^T Y$$ whose elements are the coefficients of the $N$th degree polynomial corresponding to the set of points.
Best Answer
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?