What is the y=x[p+sqrt(1+p^2)] ?

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y=x[p+sqrt(1+p^2)]

y = x [p + 1 + p^{2}]

X Intercepts : (0, 0), Y Intercepts : (0, 0)

Slope : m = p + 1 + p^{2}

Inverse : - x (p - 1 + p^{2})

Domain : - \infty < p < \infty

Range : - \infty < f (x) < \infty

Interval Notation

Domain : (- \infty, \infty)

Range : (- \infty, \infty)

Solution steps

Axis interception points of

x [p + 1 + p^{2}] :

X Intercepts

: (0, 0),

Y Intercepts

: (0, 0)

Slope of

x [p + 1 + p^{2}] : m = p + 1 + p^{2}

Inverse of

x [p + 1 + p^{2}] : - x (p - 1 + p^{2})

Domain of

x [p + 1 + p^{2}] : - \infty < p < \infty

Range of

x [p + 1 + p^{2}] : - \infty < f (x) < \infty

y = \frac{- 2}{5 x + 6}

f (x) = \frac{3}{( 1 - x ^{2} )}

f (x) = x^{5} + x^{3} - x^{2} + x + 2

y = \frac{1}{( arctan ( e ^{- 2 x} ) )}

y = (0.2 x^{3} - 0.3 x^{2} + 0.1) (- 0.1 x^{5} + 0.4 x^{2} + 4) + 0.3