What is the vertices 8x^2-48x+65 ?

The vertices 8x^2-48x+65 is Minimum (3,-7)

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Find the Vertex 8x^2-48x+65

vertices

8 x^{2} - 48 x + 65

Minimum (3, - 7)

Solution steps

y = 8 x^{2} - 48 x + 65

The parabola parameters are:

a = 8, b = - 48, c = 65

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

x_{v} = - \frac{( - 48 )}{2 \cdot 8}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 7

Therefore the parabola vertex is

(3, - 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 8

Minimum (3, - 7)

y = - 6 x^{2} + 3 x + 2

h (x) = x^{2} - 14 x + 6

x^{2} + y^{2} - 8 x - 6 y + 21 = 0

f (x) = x^{2} - 14 x + 6

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