How do you find the vertex of $y = x^{2} - 16$ $y= x^2 -16$ ?

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Answer 1 · 2018-04-29T08:49:08

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The general parabola equation

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

in this equation

$y = x^{2} - 16$ $y=x^2-16$ you have

a=1 b=0 and c=-16

$x_{v} = - \frac{b}{2 a} = - \frac{0}{2} = 0$ $x_ v=-b/(2a)=-0/2=0$

$y_{v} = {(0)}^{2} - 16 = - 16$ $y_v=(0)^2-16=-16$

the vertex is (0,-16)

graph{y=x^2-16 [-26.2, 25.1, -19.67, 6]}

How do you find the vertex of y= x^2 -16y=x2−16y= x^2 -16?