What is the vertices t^2+2t-48 ?

The vertices t^2+2t-48 is Minimum (-1,-49)

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vertices t^2+2t-48

vertices

t^{2} + 2 t - 48

Minimum (- 1, - 49)

Solution steps

y = t^{2} + 2 t - 48

The parabola parameters are:

a = 1, b = 2, c = - 48

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

t_{v} = - \frac{2}{2 \cdot 1}

Simplify

t_{v} = - 1

Plug in

t_{v} = - 1

to find the

y_{v}

value

y_{v} = - 49

Therefore the parabola vertex is

(- 1, - 49)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, - 49)

f (x) = - x^{2} + 2 x + 8

x = y^{2} - 3

y = 12 x (x - 4)

(x + 2)^{2}

y^{2} + 4 x - 2 y + 21 = 0