The problem provides a table of $x$ and $y$ values for the function $y=(x-2)^2 - a$. We are asked to find the value of $a$, draw the graph of the function, find the coordinates of the vertex, find the range of $x$ values for which the function is negative, find the larger root of the equation $(x-2)^2 - 3 = 0$, and find the value of $\sqrt{3}$ to one decimal place.

1. Problem Description

The problem provides a table of $x$ and $y$ values for the function $y=(x-2)^2 - a$. We are asked to find the value of $a$, draw the graph of the function, find the coordinates of the vertex, find the range of $x$ values for which the function is negative, find the larger root of the equation $(x-2)^2 - 3 = 0$, and find the value of $\sqrt{3}$ to one decimal place.

2. Solution Steps

(a) (i) Find the value of $a$.

We can use any pair of $(x, y)$ values from the table to find $a$. Let's use the pair $(0, 1)$.

Substituting $x = 0$ and $y = 1$ into the equation $y = (x-2)^2 - a$, we get:

$1 = (0-2)^2 - a$

$1 = (-2)^2 - a$

$1 = 4 - a$

$a = 4 - 1$

$a = 3$

(ii) Draw the graph of the function. The function is $y = (x-2)^2 - 3$.

The given values in the table are:

$x = -1, y = 6$

$x = 0, y = 1$

$x = 1, y = -2$

$x = 2, y = -3$

$x = 3, y = -2$

$x = 4, y = 1$

$x = 5, y = 6$

(b) (i) Find the coordinates of the vertex.

The vertex of the parabola is at the minimum point. From the table or from the equation $y = (x-2)^2 - 3$, we can see that the vertex is at $(2, -3)$.

(ii) Find the range of $x$ values for which the function is negative.

The function is negative when $y < 0$. From the table, we can see that $y < 0$ for $x = 1, 2, 3$. We want to find where $(x-2)^2 - 3 < 0$. This means that $(x-2)^2 < 3$, or $-\sqrt{3} < x-2 < \sqrt{3}$. So $2 - \sqrt{3} < x < 2 + \sqrt{3}$. Since $\sqrt{3} \approx 1.7$, we have $2-1.7 < x < 2+1.7$, which gives $0.3 < x < 3.7$. From the table values, $y$ is negative for $x=1$, $x=2$, and $x=3$. Therefore, the function is negative for $1 \le x \le 3$.

(iii) Find the larger root of the equation $(x-2)^2 - 3 = 0$.

$(x-2)^2 = 3$

$x-2 = \pm\sqrt{3}$

$x = 2 \pm \sqrt{3}$

The larger root is $x = 2 + \sqrt{3}$.

(iv) Find the value of $\sqrt{3}$ to one decimal place.

We know that $1^2 = 1$ and $2^2 = 4$, so $1 < \sqrt{3} < 2$.

$1.7^2 = 2.89$

$1.8^2 = 3.24$

Therefore, $1.7 < \sqrt{3} < 1.8$. Since $3$ is closer to $2.89$ than $3.24$, $\sqrt{3}$ is closer to $1.7$. Therefore $\sqrt{3} \approx 1.7$.

3. Final Answer

(a) (i) $a = 3$

(b) (i) $(2, -3)$

(ii) $2 - \sqrt{3} < x < 2 + \sqrt{3}$, which approximates to $0.3 < x < 3.7$.

(iii) $2 + \sqrt{3}$

(iv) $1.7$