What is the vertices f(x)=5x^2+1 ?

The vertices f(x)=5x^2+1 is Minimum (0,1)

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vertices f(x)=5x^2+1

vertices

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

Minimum (0, 1)

Solution steps

y = 5 x^{2} + 1

The parabola parameters are:

a = 5, b = 0, c = 1

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

x_{v} = - \frac{0}{2 \cdot 5}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(0, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 5

Minimum (0, 1)

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

x^{2} - 2 x

- x^{2} - 2 x + 3

y^{2} = 6 x - 32

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