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Popular >

f(t)=21+10t+5t^2

Solution

f (t) = 21 + 10 t + 5 t^{2}

Solution

Domain : - \infty < t < \infty

Range : f (t) \geq 16

Y Intercepts : (0, 21)

Vertex : Minimum (- 1, 16)

Inverse : \frac{- 10 + 20 t - 320}{10}, \frac{- 10 - 20 t - 320}{10}

+1

Interval Notation

Domain : (- \infty, \infty)

Range : [16, \infty)

Solution steps

Domain of

21 + 10 t + 5 t^{2} : - \infty < t < \infty

Range of

21 + 10 t + 5 t^{2} : f (t) \geq 16

Axis interception points of

21 + 10 t + 5 t^{2} :

Y Intercepts

: (0, 21)

Vertex of

21 + 10 t + 5 t^{2} :

Minimum

(- 1, 16)

Inverse of

21 + 10 t + 5 t^{2} : \frac{- 10 + 20 t - 320}{10}, \frac{- 10 - 20 t - 320}{10}

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Frequently Asked Questions (FAQ)

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