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A parabola graphed in the xy-plane has equation y=2x^(2)-5x-3. What is the x-coordinate of the vertex of the parabola?

A parabola graphed in the xy x y -plane has equation y=2x2−5x−3 y=2 x^{2}-5 x-3 . What is the x x -coordinate of the vertex of the parabola?

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Full solution

Q. A parabola graphed in the xy x y -plane has equation y=2x2−5x−3 y=2 x^{2}-5 x-3 . What is the x x -coordinate of the vertex of the parabola?
  1. Identify quadratic equation: Identify the given quadratic equation.\newlineThe given quadratic equation is y=2x2−5x−3y = 2x^2 - 5x - 3.
  2. Convert to vertex form: Convert the quadratic equation into vertex form.\newlineThe vertex form of a quadratic equation is y=a(x−h)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.\newlineTo convert the given equation into vertex form, we need to complete the square.
  3. Factor out x2x^2 coefficient: Factor out the coefficient of x2x^2 from the xx terms.\newlineThe coefficient of x2x^2 is 22. We factor out 22 from the xx terms to get:\newliney=2(x2−(5/2)x)−3y = 2(x^2 - (5/2)x) - 3
  4. Find (b/2)2(b/2)^2: Find the value of (b/2)2(b/2)^2 to complete the square.\newlineThe coefficient of xx is −5/2-5/2. We need to find (−5/4)2(-5/4)^2 to complete the square.\newline(−5/4)2=(25/16)(-5/4)^2 = (25/16)
  5. Add and subtract (b/2)2(b/2)^2: Add and subtract (b/2)2(b/2)^2 inside the parentheses and simplify.\newliney=2(x2−52x+2516)−2∗2516−3y = 2(x^2 - \frac{5}{2}x + \frac{25}{16}) - 2*\frac{25}{16} - 3\newliney=2(x−54)2−5016−3y = 2(x - \frac{5}{4})^2 - \frac{50}{16} - 3\newliney=2(x−54)2−5016−4816y = 2(x - \frac{5}{4})^2 - \frac{50}{16} - \frac{48}{16}\newliney=2(x−54)2−9816y = 2(x - \frac{5}{4})^2 - \frac{98}{16}
  6. Identify vertex x-coordinate: Identify the x-coordinate of the vertex. The x-coordinate of the vertex is the value of hh in the vertex form y=a(x−h)2+ky = a(x - h)^2 + k. From the equation y=2(x−54)2−(9816)y = 2(x - \frac{5}{4})^2 - (\frac{98}{16}), we can see that h=54h = \frac{5}{4}.

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