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Find the yy-intercept of the parabola y=x2+5x+192y = x^2 + 5x + \frac{19}{2}. \newlineSimplify your answer and write it as a proper fraction, improper fraction, or integer.\newline_____

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Q. Find the yy-intercept of the parabola y=x2+5x+192y = x^2 + 5x + \frac{19}{2}. \newlineSimplify your answer and write it as a proper fraction, improper fraction, or integer.\newline_____
  1. Evaluate at x=0x = 0: To find the y-intercept of the parabola, we need to evaluate the equation at x=0x = 0, because the y-intercept is the point where the graph of the equation crosses the y-axis, and at this point, the value of xx is always 00.
  2. Substitute x=0x = 0: Substitute x=0x = 0 into the equation y=x2+5x+192y = x^2 + 5x + \frac{19}{2} to find the y-intercept.\newliney=(0)2+5(0)+192y = (0)^2 + 5(0) + \frac{19}{2}\newliney=0+0+192y = 0 + 0 + \frac{19}{2}\newliney=192y = \frac{19}{2}

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