What is the vertices y=4x^2-40x+98 ?

The vertices y=4x^2-40x+98 is Minimum (5,-2)

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vertices y=4x^2-40x+98

vertices

y = 4 x^{2} - 40 x + 98

Minimum (5, - 2)

Solution steps

y = 4 x^{2} - 40 x + 98

The parabola parameters are:

a = 4, b = - 40, c = 98

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

x_{v} = - \frac{( - 40 )}{2 \cdot 4}

Simplify

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(5, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (5, - 2)

x = 2 y^{2}

x^{2} - 14 x + 49

y = x^{2} - 12 x + 36

f (x) = 3 x^{2} + 1

y^{2} + 2 y + x + 5 = 0

What is the vertices y=4x^2-40x+98 ?
The vertices y=4x^2-40x+98 is Minimum (5,-2)