What is the vertices 4x^2-3x+5 ?

The vertices 4x^2-3x+5 is Minimum (3/8 , 71/16)

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vertices 4x^2-3x+5

vertices

4 x^{2} - 3 x + 5

Minimum (\frac{3}{8}, \frac{71}{16})

Solution steps

y = 4 x^{2} - 3 x + 5

The parabola parameters are:

a = 4, b = - 3, c = 5

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

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

Simplify

x_{v} = \frac{3}{8}

Plug in

x_{v} = \frac{3}{8}

to find the

y_{v}

value

y_{v} = \frac{71}{16}

Therefore the parabola vertex is

(\frac{3}{8}, \frac{71}{16})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (\frac{3}{8}, \frac{71}{16})

f (x) = 5 x + 3 x^{2} - 4

f (x) = x^{2} + 40 x + 20

f (x) = - x^{2} + 4 x + 7

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

f (x) = x^{2} - 12 x + 36

What is the vertices 4x^2-3x+5 ?
The vertices 4x^2-3x+5 is Minimum (3/8 , 71/16)