What is the vertices f(x)=x^2+3x-10 ?

The vertices f(x)=x^2+3x-10 is Minimum (-3/2 ,-49/4)

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vertices f(x)=x^2+3x-10

vertices

f (x) = x^{2} + 3 x - 10

Minimum (- \frac{3}{2}, - \frac{49}{4})

Solution steps

y = x^{2} + 3 x - 10

The parabola parameters are:

a = 1, b = 3, c = - 10

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

x_{v} = - \frac{3}{2 \cdot 1}

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = - \frac{49}{4}

Therefore the parabola vertex is

(- \frac{3}{2}, - \frac{49}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{3}{2}, - \frac{49}{4})

f (x) = x^{2} + 6 x + 11

x^{2} - 6 x + 5 y = 34

f (x) = 5 x^{2} + 1

\frac{5 x ^{2}}{2}

x^{2} - 2 x

What is the vertices f(x)=x^2+3x-10 ?
The vertices f(x)=x^2+3x-10 is Minimum (-3/2 ,-49/4)