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Popular >

h(t)=-(4t^2-20t+15)

Solution

h (t) = - (4 t^{2} - 20 t + 15)

Solution

Domain : - \infty < t < \infty

Range : f (t) \leq 10

X Intercepts : (\frac{5 - 10}{2}, 0), (\frac{5 + 10}{2}, 0), Y Intercepts : (0, - 15)

Vertex : Maximum (\frac{5}{2}, 10)

Inverse : - \frac{- 5 + - t + 10}{2}, \frac{- t + 10 + 5}{2}

+1

Solution steps

Domain of

- (4 t^{2} - 20 t + 15) : - \infty < t < \infty

Range of

- (4 t^{2} - 20 t + 15) : f (t) \leq 10

Axis interception points of

- (4 t^{2} - 20 t + 15) :

X Intercepts

: (\frac{5 - 10}{2}, 0), (\frac{5 + 10}{2}, 0),

Y Intercepts

: (0, - 15)

Vertex of

- (4 t^{2} - 20 t + 15) :

Maximum

(\frac{5}{2}, 10)

Inverse of

- (4 t^{2} - 20 t + 15) : - \frac{- 5 + - t + 10}{2}, \frac{- t + 10 + 5}{2}

Popular Examples

y=(-2x+1-2)/(x+2)

y = \frac{- 2 x + 1 - 2}{x + 2}

f(x)=(x^2+4x-9)/4

f (x) = \frac{x ^{2} + 4 x - 9}{4}

y=((x^2+x-1))/((x+2))

y = \frac{( x ^{2} + x - 1 )}{( x + 2 )}

y=4+2x+3x^2+5x^3+8x^4

y = 4 + 2 x + 3 x^{2} + 5 x^{3} + 8 x^{4}

f(x)=5*(4x-1)^{-1}

f (x) = 5 \cdot (4 x - 1)^{- 1}

Frequently Asked Questions (FAQ)

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