find the vertex of f(x)=3x^2-6x+8
i have an answer of (1,-5) or is the correct answer (1,5)
To find the vertex of a quadratic function in the form f(x) = ax^2 + bx + c, you can use the formula x = -b/2a to find the x-coordinate of the vertex. Once you have the x-coordinate, you can substitute it back into the equation to find the corresponding y-coordinate.
In this case, the function is f(x) = 3x^2 - 6x + 8. Comparing it to the standard form ax^2 + bx + c, we can see that a = 3, b = -6, and c = 8.
Using the formula x = -b/2a, we substitute the values:
x = -(-6) / (2 * 3)
x = 6 / 6
x = 1
Now we have the x-coordinate of the vertex, which is x = 1. We can substitute this back into the function to find the y-coordinate:
f(1) = 3(1)^2 - 6(1) + 8
f(1) = 3(1) - 6 + 8
f(1) = 3 - 6 + 8
f(1) = 5
Therefore, the vertex of the function f(x) = 3x^2 - 6x + 8 is (1, 5). So, your answer of (1, 5) is correct. The y-coordinate of the vertex is positive 5.