What is the vertices y=k(x-0)^2+2 ?

The vertices y=k(x-0)^2+2 is {k>0: Minimum(0,2),k<0: Maximum(0,2)}

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vertices y=k(x-0)^2+2

vertices

y = k (x - 0)^{2} + 2

{k > 0 : Minimum (0, 2), k < 0 : Maximum (0, 2)}

Solution steps

y = k (x - 0)^{2} + 2

Rewrite

y = k (x - 0)^{2} + 2

in the form

y = a x^{2} + b x + c

y = k x^{2} + 2

The parabola parameters are:

a = k, b = 0, c = 2

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

x_{v} = - \frac{0}{2 k}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(0, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = k

{k > 0 : Minimum (0, 2), k < 0 : Maximum (0, 2)}

y = 5 x^{2} - 3 x - 7 + 2 x^{2} - x

4 y^{2} - 48 x - 20 y = 71

y = x^{2} + 2 x - 15

f (x) = - 4 (x + 1)^{2} - 5

x^{2} - 4

What is the vertices y=k(x-0)^2+2 ?
The vertices y=k(x-0)^2+2 is {k>0: Minimum(0,2),k<0: Maximum(0,2)}