What is the vertices f(x)=x^2-8x+15 ?

The vertices f(x)=x^2-8x+15 is Minimum (4,-1)

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vertices f(x)=x^2-8x+15

vertices

f (x) = x^{2} - 8 x + 15

Minimum (4, - 1)

Solution steps

y = x^{2} - 8 x + 15

The parabola parameters are:

a = 1, b = - 8, c = 15

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

x_{v} = - \frac{( - 8 )}{2 \cdot 1}

Simplify

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(4, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (4, - 1)

f (x) = - x^{2} + 8 x

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

y = (x + 1) (x + 2)

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

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

What is the vertices f(x)=x^2-8x+15 ?
The vertices f(x)=x^2-8x+15 is Minimum (4,-1)