We've updated our

TEXT

# Reading: Graphs of Polynomial Functions (part I)

In the previous section we explored the short run behavior of quadratics, a special case of polynomials. In this section we will explore the short run behavior of polynomials in general.

## Short run Behavior: Intercepts

As with any function, the vertical intercept can be found by evaluating the function at an input of zero. Since this is evaluation, it is relatively easy to do it for a polynomial of any degree. To find horizontal intercepts, we need to solve for when the output will be zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and 4th degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases:
1. The polynomial can be factored using known methods: greatest common factor and trinomial factoring.
2. The polynomial is given in factored form.
3. Technology is used to determine the intercepts.

### Example 1

Find the horizontal intercepts of $\displaystyle{f{{({x})}}}={x}^{{6}}-{3}{x}^{{4}}+{2}{x}^{{2}}$. We can attempt to factor this polynomial to find solutions for f(x) = 0.
 $\displaystyle{x}^{{6}}-{3}{x}^{{4}}+{2}{x}^{{2}}={0}$ Factor out the greatest common factor $\displaystyle{x}^{{2}}{({x}^{{4}}-{3}{x}^{{2}}+{2})}={0}$ Factor the inside as a quadratic in x2 $\displaystyle{x}^{{2}}{({x}^{{2}}-{1})}{({x}^{{2}}-{2})}={0}$ Break apart to find solutions $\displaystyle{x}^{{2}}={0}$ $\displaystyle{x}^{{2}}-{1}={0}$ $\displaystyle{x}^{{2}}-{2}={0}$ $\displaystyle{x}^{{2}}-{1}+{1}={0}+{1}$ $\displaystyle{x}^{{2}}-{2}+{2}={0}+{2}$ $\displaystyle{x}^{{2}}={1}$ $\displaystyle{x}^{{2}}={2}$ Our solution $\displaystyle{x}={0}$ $\displaystyle{x}=\pm\sqrt{{1}}$ $\displaystyle{x}=\pm\sqrt{{2}}$
This gives us 5 horizontal intercepts.

### Example 2

Find the vertical and horizontal intercepts of g(t) = (t – 2)2(2t + 3) The vertical intercept can be found by evaluating g(0). g(0) = (0 – 2)2(2(0) + 3) = 12 The horizontal intercepts can be found by solving g(t) = 0 ( t – 2)2(2t + 3) = 0 Since this is already factored, we can break it apart: [latex-display]\displaystyle{({t}-{2})}^{{2}}={0}[/latex-display] [latex-display]\displaystyle{t}-{2}={0}[/latex-display] [latex-display]\displaystyle{t}={2}[/latex-display] or [latex-display]\displaystyle{({2}{t}+{3})}={0}[/latex-display] [latex-display]\displaystyle{t}=\frac{{-{3}}}{{2}}[/latex-display] We can always check our answers are reasonable by graphing the polynomial.

### Example 3

Find the horizontal intercepts of h(t) = t3 + 4t2 + t – 6 Since this polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques we know, we can turn to technology to find the intercepts. Graphing this function, it appears there are horizontal intercepts at t = –3, –2, and 1. We could check these are correct by plugging in these values for t and verifying that h(–3) = h(–2) = h(1) = 0.

### Try it Now 1

Find the vertical and horizontal intercepts of the function f(t) t4 – 4t2.