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

1 Answer
Nov 30, 2017

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.