What is the vertices 4x^2-8x-3y-2=0 ?

The vertices 4x^2-8x-3y-2=0 is Minimum (1,-2)

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vertices 4x^2-8x-3y-2=0

vertices

4 x^{2} - 8 x - 3 y - 2 = 0

Minimum (1, - 2)

Solution steps

4 x^{2} - 8 x - 3 y - 2 = 0

Rewrite

4 x^{2} - 8 x - 3 y - 2 = 0

in the form

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

y = \frac{4 x ^{2}}{3} - \frac{8 x}{3} - \frac{2}{3}

The parabola parameters are:

a = \frac{4}{3}, b = - \frac{8}{3}, c = - \frac{2}{3}

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

x_{v} = - \frac{( - \frac{8}{3} )}{2 ( \frac{4}{3} )}

Simplify

- \frac{- \frac{8}{3}}{2 ( \frac{4}{3} )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(1, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{4}{3}

Minimum (1, - 2)

y = - \frac{( x + 3 ) ^{2}}{20} + 1

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

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

y = x^{2} - 5 x 3

f (x) = 2 x^{2} + 9 x + 10

What is the vertices 4x^2-8x-3y-2=0 ?
The vertices 4x^2-8x-3y-2=0 is Minimum (1,-2)