# Graph each polynomial function. f(x)=x^{4}-5x^{2}+6

Question
Graph each polynomial function. $$\displaystyle{f{{\left({x}\right)}}}={x}^{{{4}}}-{5}{x}^{{{2}}}+{6}$$

2021-02-12
Step 1
To find the x-intercepts, solve $$\displaystyle{f{{\left({x}\right)}}}={0}$$
$$\displaystyle{x}^{{{4}}}-{5}{x}^{{{2}}}+{6}={0}$$
$$\displaystyle{\left({x}^{{{2}}}-{3}\right)}{\left({x}^{{{2}}}-{2}\right)}={0}$$ Factor the trinomial
$$\displaystyle{x}^{{{2}}}-{3}={0}$$ or $$\displaystyle{x}^{{{2}}}-{2}={0}$$ Set each factor equal to 0
$$\displaystyle{x}=\pm\sqrt{{{3}}}$$ or $$\displaystyle{x}=\pm\sqrt{{{2}}}$$ Solve each equation
So, the x-intercepts are $$\displaystyle{\left(\sqrt{{{3}}},{0}\right)},{\left(\sqrt{{{2}}},{0}\right)},{\left(-\sqrt{{{3}}},{0}\right)},{\left(-\sqrt{{{2}}},{0}\right)}$$
Step 2
To find the y-intercept evaluate $$\displaystyle{f{{\left({0}\right)}}}$$
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{{4}}}-{5}{x}^{{{2}}}+{6}$$
$$\displaystyle{f{{\left({0}\right)}}}={0}^{{{4}}}-{5}{\left({0}\right)}^{{{2}}}+{6}$$ Substitute 0 for x
$$\displaystyle={6}$$ Simplify
So, the y-intercept is $$\displaystyle{\left({0},{6}\right)}$$
Plot the points (-2,2) and (2,2)

### Relevant Questions

Identify the degree of each term of the polynomial and the degree of the polynomial.
$$\displaystyle-{5}{x}^{{{3}}}+{5}{x}^{{{2}}}+{\frac{{{6}}}{{{7}{x}}}}+{4}$$
Use the theorems on derivatives to find the derivatives of the following function:
$$\displaystyle{f{{\left({x}\right)}}}={\left({5}{x}^{{{3}}}+{8}{x}^{{{2}}}-{4}\right)}^{{{4}}}{\left({8}{x}^{{{4}}}-{2}{x}^{{{3}}}-{7}\right)}$$
Use the theorems on derivatives to find the derivatives of the following function:
$$\displaystyle{f{{\left({x}\right)}}}={3}{x}^{{{5}}}-{2}{x}^{{{4}}}-{5}{x}+{7}+{4}{x}^{{-{2}}}$$
Graph each polynomial function. $$\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{{3}}}\ -\ {7}{x}^{{{2}}}\ +\ {2}{x}\ +\ {3}$$ given that 3 is a zero.
Express as a polynomial.
$$\displaystyle{\left({3}{x}+{2}\right)}{\left({x}-{5}\right)}{\left({5}{x}+{4}\right)}$$

Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q ae the same $$\displaystyle{P}{\left({x}\right)}={3}{x}^{{4}}-{5}{x}^{{3}}+{x}^{{2}}-{3}{x}+{5}\ {Q}{\left({x}\right)}={\left({\left({\left({3}{x}-{5}\right)}{x}+{1}\right)}{x}-{3}\right)}{x}+{5}$$ Try to evaluate P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polynomial $$\displaystyle{R}{\left({x}\right)}={x}^{{5}}—{2}{x}^{{4}}+{3}{x}^{{3}}—{2}{x}^{{3}}+{3}{x}+{4}$$ in “nested” form, like the polynomial Q. Use the nested form to find R(3) in your head. Do you see how calculating with the nested form follows the same arithmetic steps as calculating the value ofa polynomial using synthetic division?

Find the real zeros of $$f(x) = 5x^{3}\ -\ x^{2}\ +\ 5x\ -\ 1$$
Polynomial equation with real coefficients that has the roots $$3, 1 -i\ \text{is}\ x^{3} - 5x^{2} + 8x - 6 = 0.$$
$$\displaystyle{f{{\left({x}\right)}}}={5}{x}^{{{4}}}$$
$$\displaystyle{y}{''}={\left({4}{x}^{{{2}}}-{5}{x}+{4}\right)}{\left({3}-{x}-{2}{x}^{{{2}}}\right)},{x}={5}$$