Quadratic Functions: General Form

y = f(x) = 4x² + 3x + 2

1) Since a > 0, parabola opens upward.

2) Vertex Formula = (-b/(2a), f(-b/2a))        (version 1)
    Vertex Formula = (-b/(2a), c - b²/(2a))        (version 2)

a = 4       b = 3       c = 2

-b/(2a) = -3/8 = -3/8;
f(-b/2a) = f(-3/8) = 4(-3/8)² + 3(-3/8) + 2 = 23/16
c - b²/(4a) = 23/16

Vertex of Parabola = (-3/8, 23/16)

3) Axis of Symmetry is the vertical line x = -3/8

4) To find the x-intercept, set y = 0.

y = f(x) = 4x² + 3x + 2

0 = 4x² + 3x + 2

4x² + 3x + 2 = 0

Solutions to the equation 4x² + 3x + 2 = 0 are 0.22447894041409i and 0.22447894041409i

There are no x-intercepts since the solutions 0.22447894041409i and 0.22447894041409i are nonreal.

Parabola does not cross x-axis.

5) To find the y-intercept, set x = 0.

y = 4x² + 3x + 2

y = 4(0)² + 3(0) + 2

y = 2

Hence, y-intercept is 2.

Thus, parabola crosses y-axis at 2.

6)Domain and Range:

Domain = (-∞, ∞)

Range = [23/16, ∞)