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Popular >

h(t)=-4t^2+68t+160

Solution

h (t) = - 4 t^{2} + 68 t + 160

Solution

Domain : - \infty < t < \infty

Range : f (t) \leq 449

X Intercepts : (- \frac{- 17 + 449}{2}, 0), (\frac{17 + 449}{2}, 0), Y Intercepts : (0, 160)

Vertex : Maximum (\frac{17}{2}, 449)

Inverse : - \frac{- 17 + - t + 449}{2}, \frac{- t + 449 + 17}{2}

+1

Interval Notation

Domain : (- \infty, \infty)

Range : (- \infty, 449]

Solution steps

Domain of

- 4 t^{2} + 68 t + 160 : - \infty < t < \infty

Range of

- 4 t^{2} + 68 t + 160 : f (t) \leq 449

Axis interception points of

- 4 t^{2} + 68 t + 160 :

X Intercepts

: (- \frac{- 17 + 449}{2}, 0), (\frac{17 + 449}{2}, 0),

Y Intercepts

: (0, 160)

Vertex of

- 4 t^{2} + 68 t + 160 :

Maximum

(\frac{17}{2}, 449)

Inverse of

- 4 t^{2} + 68 t + 160 : - \frac{- 17 + - t + 449}{2}, \frac{- t + 449 + 17}{2}

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Frequently Asked Questions (FAQ)

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