# How do you find the end behavior and state the possible number of x intercepts and the value of the y intercept given y=x^4+2x^2+1?

Jan 1, 2017

${y}_{\text{intercept}} = 1$

No x-intercept.

#### Explanation:

Set $X = {x}^{2}$ giving:

$y = {X}^{2} + 2 X + 1$

$y = {\left(X + 1\right)}^{2}$

$\pm \left(X + 1\right) = \sqrt{y}$

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$\textcolor{b l u e}{\text{Determine x intercept}}$

Set $y = 0$

$- X - 1 = 0 \implies {x}^{2} = - 1$

$x = \pm \sqrt{- 1} \text{ "->" } x = \pm i$

color(green)("Thus there is no "x_("intercept"))
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$\textcolor{b l u e}{\text{Determine y intercept}}$

${y}_{\text{intercept}}$ occurs at $X = 0$

$y = {X}^{2} + 2 X + 1 \text{ "->" } y = {0}^{2} + 2 \left(0\right) + 1 = 1$

color(green)(y_("intercept")=1