Answer:
A.) The axis of symmetry:
To find the axis of symmetry, use the formula x = -b/2a, where a and b are the coefficients of the x^2 and x terms, respectively.
In this case, a = 2 and b = -12, so:
x = -(-12) / 2(2) = 3
The axis of symmetry is x = 3.
B.) The vertex:
To find the vertex, plug in the x-coordinate of the axis of symmetry (3) into the function and evaluate:
f(3) = 2(3)^2 - 12(3) + 3 = -33
So the vertex is (3, -33).
C.) The X-intercepts:
To find the x-intercepts, set y (or f(x)) equal to 0 and solve for x:
0 = 2x^2 -12x +3
Using the quadratic formula, we get:
x = (6 ± sqrt(6^2 - 4(2)(3))) / (2(2))
x = (6 ± 3sqrt(2)) / 4
x = (3/2) ± (3/2)sqrt(2)
So the x-intercepts are approximately (-0.68, 0) and (4.18, 0).
D.) The Y-intercept:
To find the y-intercept, set x = 0 and evaluate the function:
f(0) = 2(0)^2 - 12(0) + 3 = 3
So the y-intercept is (0, 3).
E.) The domain and range:
The domain of the function is all real numbers, since there are no restrictions on the values of x that can be plugged into the function.
To find the range, note that the coefficient of the x^2 term (2) is positive, which means that the parabola opens upwards. Therefore, the minimum value of the function occurs at the vertex, and the range is all real numbers greater than or equal to the y-coordinate of the vertex. In this case, the range is (-33, ∞).