Question

How do you graph `h(x)=-x^3+5x^2-7x+3`?

Answer 1

graph{-x^3+5x^2-7x+3 [-10, 10, -5, 5]}

Explanation:

This is a cubic function. The equation has degree 3, so it is an odd- degree polynomial.

The leading coefficient is negative, so as `x` approaches negative infinity, the graph rises to the left, and as `x` approaches positive infinity, the graph falls to the right.

In order to solve for the `y`-intercept, replace all the `x` variables with 0's.

`y=-(0)^3+5(0)^2-7(0)+3`

`y=3`

The constant term is 3, so the coordinate of the `y`-intercept is (0,3).

The zeroes of the function are the `x`-intercepts of the graph. In order to find these values, factor the polynomial in the equation.

List the factors of the constant term, 3: -3, -1, 1, 3

`h(1)=-(1)^3+5(1)^2-7(0)+3`

`h(1)=0`

So, `(x-1)` is a factor of `-x^3+5x^2-7x+3`.

Use synthetic division to determine the other factors.

`-x^3+5x^2-7x+3 = (x-1)(-x^2+4x-3)`

Factor the quadratic polynomial using the factor theorem.

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

`P(1)=0`

So, `(x-1)` is a factor of `-x^2+4x-3`.

Use synthetic division to determine the other factor.

`-x^3+5x^2-7x+3 = (x-1)^2(-x+3)`

The `x`-intercepts of this equation are 1 and 3.

The zero 1 has multiplicity 2, so the graph just touches the `x`-axis at the related `x`-intercept.

The zero 3 has multiplicity 1, so the graph crosses the `x`-axis at the related `x`-intercept.