+1678 Find the vertex of the graph of the equation y = -2x^2 + 8x + 15 - 3x^2 + 7x - 25
+1768 First, we combine the terms in the equation so that it is in the standard form of a quadratic equation: y=−5x2+15x−10. To find the vertex of a parabola in vertex form, we complete the square.
To complete the square, we take half of the square coefficient, square it, and add it to both sides of the equation:
y = -5x^2 + 15x - 10 y + \frac{25}{4} = -5x^2 + 15x + \frac{25}{4} - 10
The left side of the equation is now a perfect square, so we can take the square root of both sides:
y + \frac{25}{4} = \sqrt{(-5x^2 + 15x + \frac{25}{4})^2} y + \frac{25}{4} = \sqrt{(-5x + \frac{15}{2})^2}
To get the equation in vertex form, we move the constant term to the right side of the equation:
y = -\sqrt{(-5x + \frac{15}{2})^2} - \frac{25}{4}
The vertex of a parabola in vertex form is always at the point where the x -coordinate is equal to half of the coefficient of the x2 term. In this case, the coefficient of the x2 term is −5, so the vertex is at the point (1015,−425)=(23,−425).
