What is the f(x)=(sqrt(25-x^2))/(x^2-16) ?

The f(x)=(sqrt(25-x^2))/(x^2-16) is

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f(x)=(sqrt(25-x^2))/(x^2-16)

f (x) = \frac{25 - x ^{2}}{x ^{2} - 16}

Domain : - 5 \leq x < - 4 or - 4 < x < 4 or 4 < x \leq 5

Range : f (x) \leq - \frac{5}{16} or f (x) \geq 0

X Intercepts : (5, 0), (- 5, 0), Y Intercepts : (0, - \frac{5}{16})

Asymptotes : Vertical x = - 4, x = 4

Extreme Points : Minimum (- 5, 0), Maximum (0, - \frac{5}{16}), Minimum (5, 0)

Interval Notation

Domain : [- 5, - 4) \cup (- 4, 4) \cup (4, 5]

Range : (- \infty, - \frac{5}{16}] \cup [0, \infty)

Solution steps

Domain of

\frac{25 - x ^{2}}{x ^{2} - 16} : - 5 \leq x < - 4 or - 4 < x < 4 or 4 < x \leq 5

Range of

\frac{25 - x ^{2}}{x ^{2} - 16} : f (x) \leq - \frac{5}{16} or f (x) \geq 0

Axis interception points of

\frac{25 - x ^{2}}{x ^{2} - 16} :

X Intercepts

: (5, 0), (- 5, 0),

Y Intercepts

: (0, - \frac{5}{16})

Asymptotes of

\frac{25 - x ^{2}}{x ^{2} - 16} :

Vertical

: x = - 4, x = 4

Extreme Points of

\frac{25 - x ^{2}}{x ^{2} - 16} :

Minimum

(- 5, 0),

Maximum

(0, - \frac{5}{16}),

Minimum

(5, 0)

f (x) = 3 x^{2} - 4 x + 4

f (x) = x^{3} - x^{2} - x + 8

f (x, y) = sin (x)

f (x) = x^{2} + 4 x + 15

y = (x - 4)^{2} - 9