The graph of a quadratic function has a vertex at (2, -6) and passes through the po...
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Problem
The graph of a quadratic function has a vertex at (2, -6) and passes through the points (0, 2) and (4, 2). Write an equation for the function in standard form.
Step-by-step solution
The vertex form of a quadratic function is $f(x) = a(x-h)^2 + k$. Given the vertex $(h, k) = (2, -6)$, we have $f(x) = a(x-2)^2 - 6$. Using the point $(0, 2)$, we get $2 = a(0-2)^2 - 6$, which simplifies to $8 = 4a$, so $a=2$. Thus, $f(x) = 2(x-2)^2 - 6$. Expanding this to standard form $f(x) = ax^2 + bx + c$, we get $f(x) = 2(x^2 - 4x + 4) - 6 = 2x^2 - 8x + 8 - 6 = 2x^2 - 8x + 2$.
Answer
2x^2 - 8x + 2