How do you find the vertex of $y = 5 x^{2} + 4 x - 3$ $y = 5x^2 + 4x - 3$ ?

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How do you find the vertex of $y = 5 x^{2} + 4 x - 3$ $y = 5x^2 + 4x - 3$ ?

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Answer 1 · 2015-07-29T20:45:32

The vertex is at $(- \frac{2}{5}, - \frac{19}{5})$ $color(red)((-2/5,-19/5)$ .

$y = 5 x^{2} + 4 x - 3$ $y = 5x^2+4x-3$

The standard form of the equation for a parabola is

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

By comparing the two equations, we see that

$a = 5$ $a = 5$ , $b = 4$ $b = 4$ , and $c = - 3$ $c = -3$ .

The $x$ $x$ -coordinate of the vertex is at $x = - \frac{b}{2 a} = - \frac{4}{2 \times 5} = - \frac{4}{10} = - \frac{2}{5}$ $x = -b/(2a) = -4/(2×5) = -4/10 = -2/5$ .

To find the $y$ $y$ -coordinate of the vertex, we insert $x = - \frac{2}{5}$ $x = -2/5$ into the equation.

$y = 5 x^{2} + 4 x - 3 = 5 {(- \frac{2}{5})}^{2} + 4 (- \frac{2}{5}) - 3 = 5 (\frac{4}{25}) - \frac{8}{5} - 3$ $y = 5x^2+4x-3 = 5(-2/5)^2+4(-2/5)-3 = 5(4/25) -8/5-3$

$= \frac{20}{25} - \frac{8}{5} - 3 = \frac{4}{5} - \frac{8}{5} - \frac{15}{5} = - \frac{19}{5}$ $= 20/25-8/5-3 = 4/5-8/5-15/5= -19/5$

The vertex is at ( $- \frac{2}{5}, - \frac{19}{5}$ $-2/5,-19/5$ ).

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How do you find the vertex of $y = 5 x^{2} + 4 x - 3$ $y = 5x^2 + 4x - 3$ ?

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