What is the vertices f(x)=x^2-10x+24 ?

The vertices f(x)=x^2-10x+24 is Minimum (5,-1)

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vertices f(x)=x^2-10x+24

vertices

f (x) = x^{2} - 10 x + 24

Minimum (5, - 1)

Solution steps

y = x^{2} - 10 x + 24

The parabola parameters are:

a = 1, b = - 10, c = 24

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

x_{v} = - \frac{( - 10 )}{2 \cdot 1}

Simplify

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(5, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (5, - 1)

\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

f (x) = x^{2} - 8 x + 12

\frac{y ^{2}}{36} - \frac{x ^{2}}{64} = 1

\frac{( x - 3 ) ^{2}}{5} + \frac{( y - 1 ) ^{2}}{15} = 1

3 x^{2} - y^{2} = 11

What is the vertices f(x)=x^2-10x+24 ?
The vertices f(x)=x^2-10x+24 is Minimum (5,-1)