Identify the graph of a fourth degree polynomial

Looking at the graph of the polynomial given below, identify whether it is a linear/quadratic/none of these.

The point where the graph of the equation $x^2–2x–8=0$ intersects $X-$axis are $(x,0) \mathrm{and} (–2,0)$, then the value of $x$ ?

If the graph of a quadratic polynomial is a upward opening parabola, then its quadratic coefficient $a$ is .

In the above figure, the graph of a polynomial p(x) is shown. The number of zeroes of p(x) is:

If $p\left(x\right) = 4x + 5$, then the root of $p\left(x\right)$ is

The graph of $y=p\left(x\right)$ is given in the figure below, where $p\left(x\right)$ is a polynomial. Find the number of zeros.

The graph of $f\left(x\right)$ is shown below. The number of zeroes of $f\left(x\right)$ are

If the graph of a polynomial intersects the $x-\mathrm{axis}$ at only one point, it cannot be a quadratic polynomial.

The graph of the quadratic polynomial $a{x}^2+bx+c$ intersect x-axis such that $c=\frac{b^2}{4a}$, then the quadratic polynomial has

The graph of y=p(x) for the polynomial p(x) is drawn below. Find the number of zeroes of the polynomial.

Examine, seeing the graph of the polynomial given below, whether it is a linear $/$ quadratic $/$ neither linear nor quadratic.

Find the root of $p\left(x\right) = 4x + 5$

Identify the point where the graph of the equation $x^2–2x–8=0$ intersects $x-$axis.

If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial