What is the vertices-4.9t^2+9.8t ?

The vertices-4.9t^2+9.8t is Maximum (1,4.9)

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vertices-4.9t^2+9.8t

vertices

- 4.9 t^{2} + 9.8 t

Maximum (1, 4.9)

Solution steps

y = - 4.9 t^{2} + 9.8 t

The parabola parameters are:

a = - 4.9, b = 9.8, c = 0

x_{v} = - \frac{b}{2 a}

t_{v} = - \frac{9.8}{2 ( - 4.9 )}

Simplify

- \frac{9.8}{2 ( - 4.9 )} : 1

t_{v} = 1

Plug in

t_{v} = 1

to find the

y_{v}

value

y_{v} = 4.9

Therefore the parabola vertex is

(1, 4.9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 4.9

Maximum (1, 4.9)

- x^{2} + 4 x + 3

y = x^{2} - 4 x - 12

(x + 4)^{2}

f (x) = x^{2} - 6 x - 7

x^{2} - 2 x - 8