What is the vertices x^2-10x-4y+21=0 ?

The vertices x^2-10x-4y+21=0 is Minimum (5,-1)

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vertices x^2-10x-4y+21=0

vertices

x^{2} - 10 x - 4 y + 21 = 0

Minimum (5, - 1)

Solution steps

x^{2} - 10 x - 4 y + 21 = 0

Rewrite

x^{2} - 10 x - 4 y + 21 = 0

in the form

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

y = \frac{1 \cdot x ^{2}}{4} - \frac{5 x}{2} + \frac{21}{4}

The parabola parameters are:

a = \frac{1}{4}, b = - \frac{5}{2}, c = \frac{21}{4}

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

x_{v} = - \frac{( - \frac{5}{2} )}{2 ( \frac{1}{4} )}

Simplify

- \frac{- \frac{5}{2}}{2 ( \frac{1}{4} )} : 5

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(5, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{4}

Minimum (5, - 1)

f (x) = - 3 x^{2}

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

f (x) = 108 - 5 x^{2} - 7 x

2 x^{2} - x - 1

2 x^{2} - x - 6

What is the vertices x^2-10x-4y+21=0 ?
The vertices x^2-10x-4y+21=0 is Minimum (5,-1)