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

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.

$y = - {\left(0\right)}^{3} + 5 {\left(0\right)}^{2} - 7 \left(0\right) + 3$

$y = 3$

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

$h \left(1\right) = - {\left(1\right)}^{3} + 5 {\left(1\right)}^{2} - 7 \left(0\right) + 3$

$h \left(1\right) = 0$

So, $\left(x - 1\right)$ is a factor of $- {x}^{3} + 5 {x}^{2} - 7 x + 3$.

Use synthetic division to determine the other factors.

$- {x}^{3} + 5 {x}^{2} - 7 x + 3 = \left(x - 1\right) \left(- {x}^{2} + 4 x - 3\right)$

Factor the quadratic polynomial using the factor theorem.

$P \left(1\right) = - {\left(1\right)}^{2} + 4 \left(1\right) - 3$

$P \left(1\right) = 0$

So, $\left(x - 1\right)$ is a factor of $- {x}^{2} + 4 x - 3$.

Use synthetic division to determine the other factor.

$- {x}^{3} + 5 {x}^{2} - 7 x + 3 = {\left(x - 1\right)}^{2} \left(- x + 3\right)$

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.