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Popular >

v(t)=2t^2+4t+20

Solution

v (t) = 2 t^{2} + 4 t + 20

Solution

Domain : - \infty < t < \infty

Range : f (t) \geq 18

Y Intercepts : (0, 20)

Vertex : Minimum (- 1, 18)

Inverse : \frac{- 4 + 8 t - 144}{4}, \frac{- 4 - 8 t - 144}{4}

+1

Interval Notation

Domain : (- \infty, \infty)

Range : [18, \infty)

Solution steps

Domain of

2 t^{2} + 4 t + 20 : - \infty < t < \infty

Range of

2 t^{2} + 4 t + 20 : f (t) \geq 18

Axis interception points of

2 t^{2} + 4 t + 20 :

Y Intercepts

: (0, 20)

Vertex of

2 t^{2} + 4 t + 20 :

Minimum

(- 1, 18)

Inverse of

2 t^{2} + 4 t + 20 : \frac{- 4 + 8 t - 144}{4}, \frac{- 4 - 8 t - 144}{4}

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Frequently Asked Questions (FAQ)

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