What is the vertices y=-x^2+10x-28 ?

The vertices y=-x^2+10x-28 is Maximum (5,-3)

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vertices y=-x^2+10x-28

vertices

y = - x^{2} + 10 x - 28

Maximum (5, - 3)

Solution steps

y = - x^{2} + 10 x - 28

The parabola parameters are:

a = - 1, b = 10, c = - 28

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

x_{v} = - \frac{10}{2 ( - 1 )}

Simplify

- \frac{10}{2 ( - 1 )} : 5

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = - 3

Therefore the parabola vertex is

(5, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (5, - 3)

+ y^{2} = 28 x

x^{2} - 5 x + 4

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

y = 7 x^{2} - 10 x + 5

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

What is the vertices y=-x^2+10x-28 ?
The vertices y=-x^2+10x-28 is Maximum (5,-3)