Let's solve both quadratic functions by determining their standard form, vertex, domain, range, axis of symmetry, and whether the vertex is a minimum or maximum point.

1. $f(x) = 3x^2 - 6x + 1$

Standard Form:

The equation is already in the standard quadratic form, $ax^2 + bx + c$, where:

• $a = 3$
• $b = -6$
• $c = 1$

Vertex:

The vertex of a quadratic function can be found using the formula for the x-coordinate of the vertex: $x = -\frac{b}{2a}$ Substituting the values: $x = -\frac{-6}{2(3)} = 1$ To find the y-coordinate, substitute $x = 1$ back into the function: $f(1) = 3(1)^2 - 6(1) + 1 = 3 - 6 + 1 = -2$ So, the vertex is $(1, -2)$.

Domain:

The domain of any quadratic function is all real numbers: $(-\infty, \infty)$.

Range:

Since the parabola opens upwards (because $a = 3 > 0$), the vertex represents the minimum point. Thus, the range is $[-2, \infty)$.

Opening of the Graph:

The graph opens upwards because the coefficient of $x^2$ (i.e., $a = 3$) is positive.

Axis of Symmetry:

The axis of symmetry is the vertical line through the x-coordinate of the vertex: $x = 1$

Vertex Type:

Since the parabola opens upwards, the vertex $(1, -2)$ is a minimum point.

2. $f(x) = 2x^2 + 5x - 3$

Standard Form:

This is already in the standard quadratic form, where:

• $a = 2$
• $b = 5$
• $c = -3$

Vertex:

Using the formula for the x-coordinate of the vertex: $x = -\frac{b}{2a} = -\frac{5}{2(2)} = -\frac{5}{4}$ To find the y-coordinate, substitute $x = -\frac{5}{4}$ into the function: $f\left( -\frac{5}{4} \right) = 2\left( -\frac{5}{4} \right)^2 + 5\left( -\frac{5}{4} \right) - 3$ Simplifying: