What is the axis of symmetry and vertex for the graph $y = \frac{1}{20} x^{2}$ $y=1/20x^2$ ?

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What is the axis of symmetry and vertex for the graph $y = \frac{1}{20} x^{2}$ $y=1/20x^2$ ?

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Answer 1 · 2017-05-23T13:47:44

vertex: $(0, 0)$ $(0, 0)$ ; axis of symmetry: $x = 0$ $x = 0$

Explanation:

Given: $y = \frac{1}{20} x^{2}$ $y = 1/20 x^2$

Find the vertex : When $y = A x^{2} + B x + C = 0$ $y= Ax^2 + Bx + C = 0$ the vertex is $(h, k)$ $(h, k)$ , where $h = \frac{- B}{2 A}$ $h = (-B)/(2A)$ :

$h = - \frac{0}{2 \cdot \frac{1}{20}} = 0$ $h = -0/(2*1/20) = 0$

$k = f (h) = \frac{1}{20} {(0)}^{2} = 0$ $k = f(h) = 1/20 (0)^2 = 0$

$vertex : (0, 0)$ $"vertex":(0, 0)$

Find the axis of symmetry , $x = h$ $x = h$ :

axis of symmetry, $x = 0$ $x = 0$

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What is the axis of symmetry and vertex for the graph $y = \frac{1}{20} x^{2}$ $y=1/20x^2$ ?

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Explanation:

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What is the axis of symmetry and vertex for the graph $y = \frac{1}{20} x^{2}$ $y=1/20x^2$ ?