It is not that difficult to do if
You may have to move the x-axis up or down to get the maximum number of crossings. If these crossings are on the x-axis, the x-values are called the roots or zeros of the polynomial. If you have moved the axis up or down, that can be taken care of by adding or subtracting that value from the polynomial.
In the above case, if the roots are R1, R2, ... , RD then the polynomial is
y = (x - R1)*(x - R2)* ... *(x - RD)
So far so good.
Now for the tricky bits. If the root is some nasty fraction you can try to find its value by stretching the horizontal axis around that point to get a bigger scale to work from. But the root does not have to be particularly nasty: even 1/7 is difficult to make out from a graph. But if the zero is irrational you have no chance.
Next, if the polynomial has fewer than D roots then you have a different set of problems. One possibility is that the graph does cross the horizontal axes D times but some of the crossings are with the horizontal axis at one height and others are with it at another height. That means not all your factors are linear. This can happen with polynomials of degree 5 or greater.
The other possibility is that there are roots of multiplicity greater than 1. A cubic, such as y= (x-1)3 has a triple root at x = 1. If you know that it is a cubic then a single root at R simply means the polynomial is y = (x - R)3 but if you don't know the degree you are sunk. It can be very difficult to distinguish between y = ax7 and y = bx9 simply by looking at the graphs if a and b are suitable [or rather, unsuitably] chosen.
Factors
The GCF is 2.
zero
That doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (-1 plus or minus the square root of 265) divided by 12 x = 1.2732350496749756 x = -1.439901716341642
A cubic function is a smooth function (differentiable everywhere). It has no vertices anywhere.
B
a
a
graph!
Graph factor
graph apex xD
With difficulty. Plot a graph of the polynomial and see where it crosses the x axis. If it does, then y=0 at that point, and (x-a) is a factor. Sometimes you might spot where the polynomial is zero just by trying various values.
sine graph will be formed at origine of graph and cosine graph is find on y-axise
by synthetic division and quadratic equation
7421
To find the starting point of a distance vs time graph from a velocity vs time graph and a function, you would integrate the velocity function to find the displacement function. The starting point of the distance vs time graph corresponds to the initial displacement obtained from the displaced function.
If the point (x,y) is on the graph of the even function y = f(x) then so is (-x,y)