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Vertex at (5,6), and y intercept (0, -1). Write a quadratic equation?​

User Znorg
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Final answer:

The quadratic equation with a vertex at (5, 6) and y-intercept at (0, -1) is y = -7/25(x - 5)^2 + 6.

Step-by-step explanation:

To write a quadratic equation with the vertex at (5, 6) and a y-intercept at (0, -1), we can use the vertex form of a quadratic equation, which is y = a(x - h)^2 + k, where (h, k) is the vertex.

Since the vertex is at (5, 6), our equation begins with y = a(x - 5)^2 + 6. To find the value of a, we use the fact that when x = 0, the y-value is -1 (the y-intercept). Substituting these values into the equation gives -1 = a(0 - 5)^2 + 6. Simplifying this, we get -1 = 25a + 6. Solving for a, we subtract 6 from both sides, resulting in -7 = 25a, and then divide both sides by 25 to get a = -7/25.

The quadratic equation is therefore y = -7/25(x - 5)^2 + 6.

User Muge
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