Final answer:
The quadratic equation f(x) = 3x^2 + 3x - 6 can be solved for x-intercepts using the quadratic formula, yielding x = 1 and x = -2. The vertex of the parabola is (-1/2, -15/4), and the axis of symmetry is x = -1/2.
Step-by-step explanation:
To write the function f(x) = 3x^2 + 3x - 6 in intercept form and find its x-intercepts, we would need to factor the quadratic, use the quadratic formula, or complete the square. However, since this equation does not factor easily, the quadratic formula will be used to find the x-intercepts. The quadratic formula for the equation ax^2 + bx + c = 0 is x = (-b ± √(b^2 - 4ac)) / (2a).
Applying the quadratic formula to f(x) = 3x^2 + 3x - 6 where a = 3, b = 3, and c = -6 gives:
x = [-(3) ± √((3)^2 - 4(3)(-6))] / (2(3))
x = [-3 ± √(9 + 72)] / 6
x = [-3 ± √(81)] / 6
x = [-3 ± 9] / 6
So the x-intercepts are x = 1 and x = -2.
To find the vertex, we use the vertex formula h = -b / (2a) and k = f(h) with a = 3 and b = 3. So:
h = -3 / (2*3)
h = -1/2
k = 3(-1/2)^2 + 3(-1/2) - 6
k = 3(1/4) - 3/2 - 6
k = 3/4 - 3/2 - 6
k = -15/4
The vertex is thus at (-1/2, -15/4).
The axis of symmetry is given by the formula x = h, which is x = -1/2.