What is the E(x)=(x+3)^2-(2x-3)^2+(3x+1)(x-3)-9x ?

The E(x)=(x+3)^2-(2x-3)^2+(3x+1)(x-3)-9x is

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E(x)=(x+3)^2-(2x-3)^2+(3x+1)(x-3)-9x

E (x) = (x + 3)^{2} - (2 x - 3)^{2} + (3 x + 1) (x - 3) - 9 x

Domain : - \infty < x < \infty

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

X Intercepts : (3, 0), Y Intercepts : (0, - 3)

Inverse : x + 3

Interval Notation

Domain : (- \infty, \infty)

Range : (- \infty, \infty)

Solution steps

Domain of

(x + 3)^{2} - (2 x - 3)^{2} + (3 x + 1) (x - 3) - 9 x : - \infty < x < \infty

Range of

(x + 3)^{2} - (2 x - 3)^{2} + (3 x + 1) (x - 3) - 9 x : - \infty < f (x) < \infty

Axis interception points of

(x + 3)^{2} - (2 x - 3)^{2} + (3 x + 1) (x - 3) - 9 x :

X Intercepts

: (3, 0),

Y Intercepts

: (0, - 3)

Inverse of

(x + 3)^{2} - (2 x - 3)^{2} + (3 x + 1) (x - 3) - 9 x : x + 3

y = 5 \cdot (\frac{1}{2})^{- x - 1} - 2

y = 3 x^{3} - x

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

f (x) = \frac{x ^{5} + 12 x ^{3} - 5 x}{2 x ^{6} + 4 x ^{3} + x ^{2} - 2}

y = - \frac{1}{2} x^{2} + 2 x + 2.5

What is the E(x)=(x+3)^2-(2x-3)^2+(3x+1)(x-3)-9x ?
The E(x)=(x+3)^2-(2x-3)^2+(3x+1)(x-3)-9x is