What is the vertices y=4-(x-3)^2 ?

The vertices y=4-(x-3)^2 is Maximum (3,4)

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vertices y=4-(x-3)^2

vertices

y = 4 - (x - 3)^{2}

Maximum (3, 4)

Solution steps

y = 4 - (x - 3)^{2}

Rewrite

y = 4 - (x - 3)^{2}

in the form

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

y = - x^{2} + 6 x - 5

The parabola parameters are:

a = - 1, b = 6, c = - 5

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

x_{v} = - \frac{6}{2 ( - 1 )}

Simplify

- \frac{6}{2 ( - 1 )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 4

Therefore the parabola vertex is

(3, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (3, 4)

f (x) = 2 x^{2} - 10 x

f (x) = x^{2} - 6 x + 4

x^{2} + x - 4

x^{2} + x - 6

f (x) = 0.75 x^{2} - 5