What is the vertex form of $y = 2 x^{2} + 5 x - 3$ $y= 2x^2 + 5x - 3$ ?

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Answer 1 · 2016-08-13T11:48:15

$The form of equation is:$ $"The form of equation is:"$

$y = 2 {(x + \frac{5}{4})}^{2} - \frac{49}{8}$ $y=2(x+5/4)^2-49/8$

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

$y = a {(x - h)}^{2} + k Vertex form$ $y=a(x-h)^2+k" Vertex form"$

$P (h, k) represents the coordinate of vertex$ $P(h,k) "represents the coordinate of vertex"$

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

$a = 2; b = 5; c = - 3)$ $a=2" ; "b=5" ; "c=-3)$

$h = - \frac{b}{2 a}$ $h=-b/(2a)$

$h = - \frac{5}{2 \cdot 2} = - \frac{5}{4}$ $h=-5/(2*2)=-5/4$

$k = 2 \cdot {(- \frac{5}{4})}^{2} + 5 \cdot (- \frac{5}{4}) - 3$ $k=2*(-5/4)^2+5*(-5/4)-3$

$k = 2 \cdot \frac{25}{16} - \frac{25}{4} - 3$ $k=2*25/16-25/4-3$

$k = \frac{50}{16} - \frac{25}{4} - 3$ $k=50/16-25/4-3$

$k = \frac{50 - 100 - 48}{16}$ $k=(50-100-48)/16$

$k = - \frac{49}{8} = - 6.13 Rounded the two decimal place$ $k=-49/8=-6.13" Rounded the two decimal place"$

$The form of equation is:$ $"The form of equation is:"$

$y = 2 {(x + \frac{5}{4})}^{2} - \frac{49}{8}$ $y=2(x+5/4)^2-49/8$

What is the vertex form of y= 2x^2 + 5x - 3 y=2x2+5x−3y= 2x^2 + 5x - 3 ?