What is the f(x)=-1/40 x^2+7/4 x,0<= x<= 120 ?

The f(x)=-1/40 x^2+7/4 x,0<= x<= 120 is

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f(x)=-1/40 x^2+7/4 x,0<= x<= 120

f (x) = - \frac{1}{40} x^{2} + \frac{7}{4} x, 0 \leq x \leq 120

Domain : 0 \leq x \leq 120

Range : - 150 \leq f (x) \leq \frac{245}{8}

X Intercepts : (0, 0), (70, 0), Y Intercepts : (0, 0)

Vertex : Maximum (35, \frac{245}{8})

Inverse : - \frac{- 70 + 4900 - 160 x}{2}, - \frac{- 70 - 4900 - 160 x}{2}

Interval Notation

Domain : [0, 120]

Range : [- 150, \frac{245}{8}]

Solution steps

Domain of

- \frac{1}{40} x^{2} + \frac{7}{4} x, 0 \leq x \leq 120 : 0 \leq x \leq 120

Range of

- \frac{1}{40} x^{2} + \frac{7}{4} x, 0 \leq x \leq 120 : - 150 \leq f (x) \leq \frac{245}{8}

Axis interception points of

- \frac{1}{40} x^{2} + \frac{7}{4} x, 0 \leq x \leq 120 :

X Intercepts

: (0, 0), (70, 0),

Y Intercepts

: (0, 0)

Vertex of

- \frac{1}{40} x^{2} + \frac{7}{4} x, 0 \leq x \leq 120 :

Maximum

(35, \frac{245}{8})

Inverse of

- \frac{1}{40} x^{2} + \frac{7}{4} x, 0 \leq x \leq 120 : - \frac{- 70 + 4900 - 160 x}{2}, - \frac{- 70 - 4900 - 160 x}{2}

f (m) = 1 - m

f (x) = e^{x}

g (x) = 3 x^{5} - 20 x^{3}

f (x) = \frac{( x - 6 ) 3 x - 9}{2 x ^{2} - 72}

y = - \frac{1}{12} x^{2} + 2 x + 3

What is the f(x)=-1/40 x^2+7/4 x,0<= x<= 120 ?
The f(x)=-1/40 x^2+7/4 x,0<= x<= 120 is