Question

What are the standard form and the factored form of the function?

Answer 1

Answer:

Factored form: $P(x)=-\frac{1}{1500}(x+20)(x+5)(x-15)$.

Standard form: $P(x)=-\frac{x^3}{1500}-\frac{x^2}{150}+\frac{11x}{60}+1$

Step-by-step explanation:

The factor form of a polynomial is

$P(x)=a(x-c_1)(x-c_2)...(x-c_n)$

where, a is a constant and leading coefficient, $c_1,c_2, ...,c_n$ are n zeroes of the polynomial.

From the given graph it is clear that the x-intercepts of the function are -20, -5 and 15. It means the zeroes of the given function are -20, -5 and 15. So, the required function is

$P(x)=a(x-(-20))(x-(-5))(x-15)$

$P(x)=a(x+20)(x+5)(x-15)$           .... (1)

From the given graph it is clear that y-intercept of the function is (0,1). Use the y-intercept of the graph to find the value of a.

$1=a(0+20)(0+5)(0-15)$

$1=-1500a$

Divide both sides by -1500.

$\frac{1}{-1500}=a$

$-\frac{1}{1500}=a$

Put $a=-\frac{1}{1500}$ in equation (1).

$P(x)=-\frac{1}{1500}(x+20)(x+5)(x-15)$

Therefore the factored form of the function is $P(x)=-\frac{1}{1500}(x+20)(x+5)(x-15)$.

Expand the above function to find the standard form of the function

$P(x) = \frac{-x^3 - 10 x^2 + 275 x + 1500}{1500}$

$P(x)=-\frac{x^3}{1500}-\frac{x^2}{150}+\frac{11x}{60}+1$

Therefore the standard form of the function is $P(x)=-\frac{x^3}{1500}-\frac{x^2}{150}+\frac{11x}{60}+1$.

Answer 2

Answer:

f(x)= -(x^3+10x^2-275x-1500)  Standard

f(x)= -(x+20)(x+5)(x-15)

Step-by-step explanation:

The factored form of the equation is:

(x+20)(x+5)(x-15)=f(x)  Each of the zeroes are where the graph crosses the x-axis.

The expanded form of the equation is found by using FOIL on the equations above:

-x^3+10x^2-275X-1500

The leading coefficient is negative because the graph rises to the left and falls to the right.