How do you find the y coordinate of the vertex of a parabola?

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Answer 1 · 2018-07-07T06:11:41

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Given: a parabola equation

If the parabola equation is in the form: $f (x) = A x^{2} + B x + C = 0$ $f(x) = Ax^2 + Bx + C = 0$

The vertex is $(- \frac{B}{2 A}, f (- \frac{B}{2 A}))$ $(-B/(2A), f(-B/(2A)))$

Once you have the $x$ $x$ value of the vertex, just evaluate the function with that $x$ $x$ -value.

Example: $f (x) = 3 x^{2} - 2 x - 9$ $f(x) = 3x^2 -2x - 9$

$x = - \frac{- 2}{2 \cdot 3} = \frac{2}{6} = \frac{1}{3}$ $x = -(-2)/(2*3) = 2/6 = 1/3$

$f (\frac{1}{3}) = 3 {(\frac{1}{3})}^{2} - \frac{2}{1} \cdot \frac{1}{3} - 9$ $f(1/3) = 3 (1/3)^2 - 2/1 * 1/3 - 9$

$f (\frac{1}{3}) = \frac{3}{1} \cdot \frac{1}{9} - \frac{2}{3} - 9$ $f(1/3) = 3/1 * 1/9 - 2/3 - 9$

$f (\frac{1}{3}) = \frac{3}{9} - \frac{2}{3} - 9$ $f(1/3) = 3/9 - 2/3 - 9$

$f (\frac{1}{3}) = \frac{1}{3} - \frac{2}{3} - \frac{9}{1} \cdot \frac{3}{3}$ $f(1/3) = 1/3 - 2/3 - 9/1 * 3/3$

$f (\frac{1}{3}) = - \frac{1}{3} - \frac{27}{3} = - \frac{28}{3} = - 9 \frac{1}{3}$ $f(1/3) = -1/3 - 27/3 = -28/3 = -9 1/3$

vertex: $(\frac{1}{3}, - \frac{28}{3})$ $(1/3, -28/3)$