To find the turning point of a parabola in the form of f(x) = ax^2 + bx + c, first complete the square to rewrite the equation in vertex form.
Given f(x) = −3x^2 + 12x − 4, we complete the square as follows:
f(x) = -3(x^2 - 4x) - 4
f(x) = -3(x^2 - 4x + 4) - 4 + 12
f(x) = -3(x - 2)^2 + 8
Thus, the vertex form of the equation is h = 2, and k = 8.
Therefore, the turning point of the parabola is when x = 2, and y = 8.
Determine the turning point of the parabola f(x)=−3x2+12x−4
1 answer