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Popular >

h(t)=-t^2+3t+10

Solution

h (t) = - t^{2} + 3 t + 10

Solution

Domain : - \infty < t < \infty

Range : f (t) \leq \frac{49}{4}

X Intercepts : (- 2, 0), (5, 0), Y Intercepts : (0, 10)

Vertex : Maximum (\frac{3}{2}, \frac{49}{4})

Inverse : - \frac{- 3 + - 4 t + 49}{2}, - \frac{- 3 - - 4 t + 49}{2}

+1

Interval Notation

Domain : (- \infty, \infty)

Range : (- \infty, \frac{49}{4}]

Solution steps

Domain of

- t^{2} + 3 t + 10 : - \infty < t < \infty

Range of

- t^{2} + 3 t + 10 : f (t) \leq \frac{49}{4}

Axis interception points of

- t^{2} + 3 t + 10 :

X Intercepts

: (- 2, 0), (5, 0),

Y Intercepts

: (0, 10)

Vertex of

- t^{2} + 3 t + 10 :

Maximum

(\frac{3}{2}, \frac{49}{4})

Inverse of

- t^{2} + 3 t + 10 : - \frac{- 3 + - 4 t + 49}{2}, - \frac{- 3 - - 4 t + 49}{2}

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Frequently Asked Questions (FAQ)

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