The problem asks to graph the given quadratic functions and find the vertex and axis of symmetry for each. The given functions are: (1) $y = x^2 - 6x + 5$ (2) $y = 2x^2 + 8x + 3$ (3) $y = -3x^2 + 6x + 1$ (4) $y = -x^2 - 4x + 2$

1. Problem Description

The problem asks to graph the given quadratic functions and find the vertex and axis of symmetry for each. The given functions are:

(1) $y = x^2 - 6x + 5$

(2) $y = 2x^2 + 8x + 3$

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

(4) $y = -x^2 - 4x + 2$

2. Solution Steps

We will complete the square for each quadratic function to find the vertex form, $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex and $x = h$ is the axis of symmetry.

(1) $y = x^2 - 6x + 5$

$y = (x^2 - 6x + 9) - 9 + 5$

$y = (x - 3)^2 - 4$

Vertex: $(3, -4)$

Axis of symmetry: $x = 3$

(2) $y = 2x^2 + 8x + 3$

$y = 2(x^2 + 4x) + 3$

$y = 2(x^2 + 4x + 4) - 2(4) + 3$

$y = 2(x + 2)^2 - 8 + 3$

$y = 2(x + 2)^2 - 5$

Vertex: $(-2, -5)$

Axis of symmetry: $x = -2$

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

$y = -3(x^2 - 2x) + 1$

$y = -3(x^2 - 2x + 1) -3(-1) + 1$

$y = -3(x - 1)^2 + 3 + 1$

$y = -3(x - 1)^2 + 4$

Vertex: $(1, 4)$

Axis of symmetry: $x = 1$

(4) $y = -x^2 - 4x + 2$

$y = -(x^2 + 4x) + 2$

$y = -(x^2 + 4x + 4) -(-4) + 2$

$y = -(x + 2)^2 + 4 + 2$

$y = -(x + 2)^2 + 6$

Vertex: $(-2, 6)$

Axis of symmetry: $x = -2$

3. Final Answer

(1) Vertex: $(3, -4)$, Axis of symmetry: $x = 3$

(2) Vertex: $(-2, -5)$, Axis of symmetry: $x = -2$

(3) Vertex: $(1, 4)$, Axis of symmetry: $x = 1$

(4) Vertex: $(-2, 6)$, Axis of symmetry: $x = -2$