# How do you use end behavior, zeros, y intercepts to sketch the graph of g(x)=(x+1)(x-2)^2(x-4)?

Nov 29, 2017

graph{(x+1)(x-2)^2(x-4) [-10, 10, -5, 5]}

#### Explanation:

This is a quartic function. The equation has degree 4, so it is an even- degree polynomial.

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

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

$y = \left(0 + 1\right) {\left(0 - 2\right)}^{2} \left(0 - 4\right)$

$y = \left(1\right) \left(4\right) \left(- 4\right)$

$y = - 16$

The constant term is -16, so the coordinate of the $y$-intercept is (0,-16).

The zeroes of the function are the $x$-intercepts of the graph. The $x$-intercepts of this equation are -1, 2, and 4.

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

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

The zero 4 has multiplicity 1, so the graph crosses the $x$-axis at the related $x$-intercept.