The x-intercepts of the graphed polynomial are

(0,0), (-1, 0) and (3, 0), as can be seen from the graph. These are the points where the curve intersects x-axis.

The y-intercept of the polynomial is (0, 0), where the curve intersects y-axis.

The x-intercepts of the polynomial are the points...

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The x-intercepts of the graphed polynomial are

(0,0), (-1, 0) and (3, 0), as can be seen from the graph. These are the points where the curve intersects x-axis.

The y-intercept of the polynomial is (0, 0), where the curve intersects y-axis.

The x-intercepts of the polynomial are the points where the polynomial is zero:

f(x) = 0. Note that at the points x = -1 and x = 3 the curve crosses x-axis, so the value of the polynomial f(x) changes its sign (from positive to negative, or vice versa). This means x - 3 and x + 1 are the factors of the polynomial.

However, at x =0, the curve just touches x-axis and the polynomial does not change its sign. This means that this is a zero *of the second order*, which indicates that one of the factors of the polynomial is x^2.

Thus, this polynomial can be written in factored form as

f(x) = ax^2 (x - 3)(x + 1), where a is the leading coefficient.

We can find a by using the fact that the graph goes through point (5, -60). For x = 5, f(5) = -60:

f(5) = a*5^2 * (5 - 3)(5 + 1) = -60

25a * 2 * 6 = -60

Dividing both sides of this equation by 2, 6 and 5, we get

5a = -1

So a = -1/5 = -0.2, and f(x) = -0.2x^2 (x -3)(x + 1).

Multiplying it out, we get

f(x) = -0.2x^2(x^2 - 2x - 3) = -0.2x^4 + 0.4x^3 + 0.6x^2.

To summarize:

**x-intercepts are (3,0) and (-1, 0). (0,0) is both x and y intercept.**

**Factors are x^2, (x - 3) and (x + 1).**

**The leading coefficient is a = -0.2.**

**The polynomial is f(x) = -0.2x^4 + 0.4x^3 + 0.6x^2.**

Intercepts are easily read from the graph, those are the points where graph of the function goes through or touches axes.

y-intercept is 0.

x-intercepts are -1, 0 and 3. However we must bear in mind that intercept at point 0 is different than the other two because at that point function doesn't go through x axis rather it only touches it. Therefore, `x=0` will be root of multiplicity 2.

If we have all roots (x-intercepts) `x_1,x_2,ldots,x_n` of the polynomial `P,` then we can write the polynomial as product of its factors `P(x)=(x-x_1)(x-x_2)cdots(x-x_n)`

In your case factors would be `x,` `x+1` and `x-3` . Now we can write our polynomial

`P(x)=ax^2(x+1)(x-3)`

`a` is unknown coefficient. `x` is squared because, as we said earlier 0 is root of multiplicity 2 (hence the power 2).

Now to find `a` we simply plug in the point through which the graph passes (5,-60).

`P(5)=-60`

`a cdot 5^2(5+1)(5-3)=-60`

`300a=-60`

`a=-60/300`

`a=-1/5`

So your polynomial is

`P(x)=-1/5x^2(x+1)(x-3)`

`P(x)=-1/5x^4+2/5 x^3+3/5 x^2`