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Popular >

h(t)=-16t^2+128t+50

Solution

h (t) = - 16 t^{2} + 128 t + 50

Solution

Domain : - \infty < t < \infty

Range : f (t) \leq 306

X Intercepts : (- \frac{- 16 + 3 34}{4}, 0), (\frac{16 + 3 34}{4}, 0), Y Intercepts : (0, 50)

Vertex : Maximum (4, 306)

Inverse : - \frac{- 16 + - t + 306}{4}, \frac{- t + 306 + 16}{4}

+1

Interval Notation

Domain : (- \infty, \infty)

Range : (- \infty, 306]

Solution steps

Domain of

- 16 t^{2} + 128 t + 50 : - \infty < t < \infty

Range of

- 16 t^{2} + 128 t + 50 : f (t) \leq 306

Axis interception points of

- 16 t^{2} + 128 t + 50 :

X Intercepts

: (- \frac{- 16 + 3 34}{4}, 0), (\frac{16 + 3 34}{4}, 0),

Y Intercepts

: (0, 50)

Vertex of

- 16 t^{2} + 128 t + 50 :

Maximum

(4, 306)

Inverse of

- 16 t^{2} + 128 t + 50 : - \frac{- 16 + - t + 306}{4}, \frac{- t + 306 + 16}{4}

Popular Examples

f(x)=(log_{6}(x))/(log_{7)(x)}

f (x) = \frac{lo g _{6} ( x )}{lo g _{7} ( x )}

f (x) = 1 - sec (x)

f(t)=3t^{5/2}-cos(3t)

f (t) = 3 t^{\frac{5}{2}} - cos (3 t)

f(x)=-1/4 sqrt(x-5)+9

f (x) = - \frac{1}{4} x - 5 + 9

y=log_{1/2}(x-1)

y = lo g_{\frac{1}{2}} (x - 1)

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