# How do you find intercepts, extrema, points of inflections, asymptotes and graph y=xsqrt(16-x^2)?

Jan 11, 2017

$f \left(x\right) = x \sqrt{16 - {x}^{2}}$ is defined for $x \in \left[- 4 , 4\right]$, has no asymptotes, intercepts $y = 0$ at ${x}_{1} = - 4$, ${x}_{2} = 0$, ${x}_{3} = 4$, has an absolute minimum for ${x}_{m} = - 2 \sqrt{2}$ and absolute maximum at ${x}_{M} = 2 \sqrt{2}$ and an inflection point for $x = 0$.

#### Explanation:

Consider:

$f \left(x\right) = x \sqrt{16 - {x}^{2}}$

The function is defined for $16 - {x}^{2} \ge 0$ that is for $x \in \left[- 4 , 4\right]$ and it is continuous in its domain.

At the function is continuous in a compact domain, it cannot have asymptotes.

(1) We can find the intercepts solving the equation:

$x \sqrt{16 - {x}^{2}} = 0$

which yields three roots:

${x}_{1} = - 4$, ${x}_{2} = 0$, ${x}_{3} = 4$

(2) Calculate the first derivative:

$f ' \left(x\right) = \sqrt{16 - {x}^{2}} - {x}^{2} / \sqrt{16 - {x}^{2}} = \frac{16 - {x}^{2} - {x}^{2}}{\sqrt{16 - {x}^{2}}} = \frac{16 - 2 {x}^{2}}{\sqrt{16 - {x}^{2}}}$

so $f \left(x\right)$ is only differentiable in the open interval $x \in \left(- 4 , 4\right)$, and the critical points can be determined as:

$f ' \left(x\right) = 0$

$\frac{16 - 2 {x}^{2}}{\sqrt{16 - {x}^{2}}} = 0$

$16 - 2 {x}^{2} = 0$

$x = \pm 2 \sqrt{2}$

Solving the inequality:

$f ' \left(x\right) > 0$

we can see that:

$\frac{16 - 2 {x}^{2}}{\sqrt{16 - {x}^{2}}} > 0$ for $\left\mid x \right\mid < 2 \sqrt{2}$

This means that $f \left(x\right)$ is:

• decreasing in the interval $x \in \left[- 4 , - 2 \sqrt{2}\right]$

• increasing in the interval $x \in \left[- 2 \sqrt{2} , 2 \sqrt{2}\right]$

• decreasing in the interval $x \in \left[2 \sqrt{2} , 4\right]$

and thus:

• ${x}_{m} = - 2 \sqrt{2}$ is a local minimum
• ${x}_{M} = 2 \sqrt{2}$ is a local maximum

Based on Weierstrass' Theorem, as $f \left(x\right)$ is continuous in a compact domain it also has absolute minimum and maximum, and we can easily check that ${x}_{m}$ and ${x}_{M}$ are the absolute minimum and maximum respectively.

(3) We calculate the second derivative:

$f ' ' \left(x\right) = \frac{- 4 x \sqrt{16 - {x}^{2}} + x \frac{16 - 2 {x}^{2}}{\sqrt{16 - {x}^{2}}}}{16 - {x}^{2}} = \frac{- 4 x \left(16 - {x}^{2}\right) + x \left(16 - 2 {x}^{2}\right)}{{\left(16 - {x}^{2}\right)}^{\frac{3}{2}}} = \frac{x \left(- 64 + 4 {x}^{2} + 16 - 2 {x}^{2}\right)}{{\left(16 - {x}^{2}\right)}^{\frac{3}{2}}} = \frac{2 x \left({x}^{2} - 24\right)}{{\left(16 - {x}^{2}\right)}^{\frac{3}{2}}}$

and we can see that $f \left(x\right)$has only one inflection point in its domain at $x = 0$, since ${x}^{2} - 24 < 0$ for $x \in \left(- 4 , 4\right)$, so that:

• $f ' ' \left(x\right) > 0$ for $x \in \left(- 4 , 0\right)$ and so $f \left(x\right)$ is concave up.
• $f ' ' \left(x\right) < 0$ for $x \in \left(0 , 4\right)$ and so $f \left(x\right)$ is concave down.

graph{x sqrt(16-x^2) [-20, 20, -10, 10]}

Jan 11, 2017

See explanation.

#### Explanation:

To make y real, $x \in \left[- 4 , 4\right]$

x-intercept ( y = 0 ) : $\pm 4$

The graph also passes through the origin

Upon substitution

$x = 4 \sin \theta$,

$y = 16 \sin \theta \cos \theta = 8 \sin 2 \theta$, revealing that

$y \in \left[- 8 , 8\right] .$.

The graph is contained in an 8$\times$16 rectangle, and so, the

question of search for asymptote(s) does not arise.

$y ' = \frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}} = \frac{16 \cos 2 \theta}{4 \cos \theta}$

$= 4 \cos 2 \theta \sec \theta$

=0, at $\theta = \pm \frac{\pi}{4} \to x = \pm 2 \sqrt{2} = \pm 2.83$, nearly.

Max y = 8, at $\theta = \frac{\pi}{4}$

Min y = $- 8$, at $\theta = - \frac{\pi}{4}$.

y''=((dy')/(d theta))/((dx)/(d theta)

$= 4 \left(\left(- 2 \sin 2 \theta \sec \theta + \cos 2 \theta \sec \theta \tan \theta\right)\right)$

$/ \left(4 \cos \theta\right)$

=0, only when $\theta = 0$ ( after simplification, sin theta is a factor in

the numerator ).

The origin is the point of inflexion ( after studying tangent-crossing -

curve, at $\theta = 0$.

graph{(y-xsqrt(16-x^2))(y-8.1)(y+8.1)=0[-5, 5, -9, 9]}