Tap the blue points to see coordinates.
STEP 1:Find the x-intercepts
To find the x-intercepts solve, the equation $ \color{blue}{ -x^4+2x^2+x = 0 } $
The solutions of this equation are:
$$ \begin{matrix}x_1 = 0 & x_2 = -1 & x_3 = 1.618 & x_4 = -0.618 \end{matrix} $$(you can use the step-by-step polynomial equation solver to see a detailed explanation of how to solve the equation)
STEP 2:Find the y-intercepts
To find the y-intercepts, substitute $ x = 0 $ into $ \color{blue}{ p(x) = -x^4+2x^2+x } $, so:
$$ \text{Y inercept} = p(0) = 0 $$STEP 3:Find the end behavior
The end behavior of a polynomial is the same as the end behavior of a leading term.
$$ \lim_{x \to -\infty} \left( -x^4+2x^2+x \right) = \lim_{x \to -\infty} -x^4 = \color{blue}{ -\infty } $$The graph starts in the lower-left corner.
$$ \lim_{x \to \infty} \left( -x^4+2x^2+x \right) = \lim_{x \to \infty} -x^4 = \color{blue}{ -\infty } $$The graph ends in the lower-right corner.
STEP 4:Find the turning points
To determine the turning points, we need to find the first derivative of $ p(x) $:
$$ p^{\prime} (x) = -4x^3+4x+1 $$The x coordinate of the turning points are located at the zeros of the first derivative
$$ p^{\prime} (x) = 0 $$ $$ \begin{matrix}x_1 = -0.2696 & x_2 = -0.8376 & x_3 = 1.1072 \end{matrix} $$(cleck here to see a explanation of how to solve the equation)
To find the y coordinates, substitute the above values into $ p(x) $
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ -0.2696 } \Rightarrow p\left(-0.2696\right) = \color{orangered}{ -0.1295 }\\[1 em] \text{for } ~ x & = \color{blue}{ -0.8376 } \Rightarrow p\left(-0.8376\right) = \color{orangered}{ 0.0733 }\\[1 em] \text{for } ~ x & = \color{blue}{ 1.1072 } \Rightarrow p\left(1.1072\right) = \color{orangered}{ 2.0562 }\end{aligned} $$So the turning points are:
$$ \begin{matrix} \left( -0.2696, -0.1295 \right) & \left( -0.8376, 0.0733 \right) & \left( 1.1072, 2.0562 \right)\end{matrix} $$STEP 5:Find the inflection points
The inflection points are located at zeroes of second derivative. The second derivative is $ p^{\prime \prime} (x) = -12x^2+4 $.
The zeros of second derivative are
$$ \begin{matrix}x_1 = \dfrac{\sqrt{ 3 }}{ 3 } & x_2 = - \dfrac{\sqrt{ 3 }}{ 3 } \end{matrix} $$Substitute the x values into $ p(x) $ to get y coordinates
$$ \begin{aligned} \text{for } ~ x & = \color{blue}{ \frac{\sqrt{ 3 }}{ 3 } } \Rightarrow p\left(\frac{\sqrt{ 3 }}{ 3 }\right) = \color{orangered}{ 1.1329 }\\[1 em] \text{for } ~ x & = \color{blue}{ - \frac{\sqrt{ 3 }}{ 3 } } \Rightarrow p\left(- \frac{\sqrt{ 3 }}{ 3 }\right) = \color{orangered}{ -0.0218 }\end{aligned} $$So the inflection points are:
$$ \begin{matrix} \left( \dfrac{\sqrt{ 3 }}{ 3 }, 1.1329 \right) & \left( - \dfrac{\sqrt{ 3 }}{ 3 }, -0.0218 \right)\end{matrix} $$