What is the vertices 2x^2+3x-20 ?

The vertices 2x^2+3x-20 is Minimum (-3/4 ,-169/8)

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vertices 2x^2+3x-20

vertices

2 x^{2} + 3 x - 20

Minimum (- \frac{3}{4}, - \frac{169}{8})

Solution steps

y = 2 x^{2} + 3 x - 20

The parabola parameters are:

a = 2, b = 3, c = - 20

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

x_{v} = - \frac{3}{2 \cdot 2}

Simplify

x_{v} = - \frac{3}{4}

Plug in

x_{v} = - \frac{3}{4}

to find the

y_{v}

value

y_{v} = - \frac{169}{8}

Therefore the parabola vertex is

(- \frac{3}{4}, - \frac{169}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- \frac{3}{4}, - \frac{169}{8})

y = x^{2} - 2 x^{1}

y = x^{2} - 2 x^{2}

y = 2 x^{2} + 3

y = 2 x^{2} - 3

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

What is the vertices 2x^2+3x-20 ?
The vertices 2x^2+3x-20 is Minimum (-3/4 ,-169/8)