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

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.