The problem consists of four parts. Part 1: Given the function $y = (2+x)(x-4)$, we need to sketch the graph of the function and find the minimum value of $y$. Part 2: We need to factorize the equation $2x^2 + 5x - 3$ completely. Part 3: Given the graph of the function $y = 3 - 2x - x^2$, we need to find the coordinates of the points A and C.

1. Problem Description

The problem consists of four parts.

Part 1: Given the function $y = (2+x)(x-4)$, we need to sketch the graph of the function and find the minimum value of $y$.

Part 2: We need to factorize the equation $2x^2 + 5x - 3$ completely.

Part 3: Given the graph of the function $y = 3 - 2x - x^2$, we need to find the coordinates of the points A and C.

2. Solution Steps

Part 1:

i. Sketch the graph of $y = (2+x)(x-4)$

Expanding the equation, we have $y = x^2 - 2x - 8$.

This is a parabola. The roots are $x = -2$ and $x = 4$.

The x-coordinate of the vertex is the average of the roots: $x_v = \frac{-2 + 4}{2} = 1$.

The y-coordinate of the vertex is found by substituting $x_v$ into the equation: $y_v = (2+1)(1-4) = 3(-3) = -9$.

The vertex is at $(1, -9)$.

The y-intercept is when $x = 0$, so $y = (2+0)(0-4) = -8$.

Since the coefficient of $x^2$ is positive, the parabola opens upwards.

ii. Find the minimum value of $y$.

The minimum value of $y$ occurs at the vertex of the parabola.

The y-coordinate of the vertex is $-9$. Therefore, the minimum value of $y$ is $-9$.

Part 2:

Factorize the equation $2x^2 + 5x - 3$.

We are looking for two numbers that multiply to $(2)(-3) = -6$ and add up to $5$. The numbers are $6$ and $-1$.

Rewrite the middle term: $2x^2 + 6x - x - 3$.

Factor by grouping: $2x(x+3) - 1(x+3)$.

Factor out the common term $(x+3)$: $(2x - 1)(x+3)$.

Part 3:

Find the coordinates of the points A and C.

The points A and C are the x-intercepts of the graph $y = 3 - 2x - x^2$.

To find the x-intercepts, set $y = 0$: $0 = 3 - 2x - x^2$.

Multiply by $-1$: $x^2 + 2x - 3 = 0$.

Factor the quadratic equation: $(x+3)(x-1) = 0$.

The roots are $x = -3$ and $x = 1$.

Since point A is to the left of the y-axis and C is to the right, $A = (-3, 0)$ and $C = (1, 0)$.

3. Final Answer

Part 1:

i. Sketch: A parabola with roots at $x=-2$ and $x=4$, vertex at $(1,-9)$ and y-intercept at $y=-8$.

ii. Minimum value of $y = -9$

Part 2:

$(2x - 1)(x+3)$

Part 3:

A = $(-3, 0)$ and C = $(1, 0)$