The problem provides a table of $x$ and $y$ values, where $y$ is a quadratic function of $x$. The table has some missing $y$ values. We are asked to find the missing $y$ value when $x = 3$, sketch the graph, find the axis of symmetry, find the turning point, the range of $x$ where $y$ is positive, and express the quadratic function in the form $y = -x^2 + ax + b$, then determine the values of $a$ and $b$.

1. Problem Description

The problem provides a table of $x$ and $y$ values, where $y$ is a quadratic function of $x$. The table has some missing $y$ values. We are asked to find the missing $y$ value when $x = 3$, sketch the graph, find the axis of symmetry, find the turning point, the range of $x$ where $y$ is positive, and express the quadratic function in the form $y = -x^2 + ax + b$, then determine the values of $a$ and $b$.

2. Solution Steps

(i) Finding the missing value of $y$ when $x=3$.

We observe the table:

x | -1 | 0 | 1 | 2 | 3 | 4 | 5

---|---|---|---|---|---|---|---

y | -4 | 1 | 4 | 5 | ? | 1 | -4

The graph is symmetric. The axis of symmetry should be in the middle. We see that $y$ values are the same when $x = -1$ and $x = 5$, i.e., -

4. Also, when $x=0$ and $x=4$, the $y$ value is $1$. This indicates that the axis of symmetry is at $x = 2$. The values of $x$ are symmetric around $x = 2$.

$x = -1, 0, 1, 2, 3, 4, 5$ are

$-1, 0, 1$ are $-3, -2, -1$ from 2,

$3, 4, 5$ are $1, 2, 3$ from

2.

Thus, $x=3$ corresponds to $x=1$, so $y=4$.

(ii) Sketching the graph. Since the coefficient of $x^2$ is negative, the parabola opens downwards.

(iii) (a) The axis of symmetry is $x=2$.

(b) Turning point: The vertex is the maximum point of the parabola. From the table and by understanding that the axis of symmetry is at $x=2$, we see that when $x=2$, $y=5$. Thus, the vertex is $(2,5)$.

(iv) To find when $y$ is positive, we look at the table. $y > 0$ when $x=0, 1, 2, 3, 4$. Therefore the interval where $y$ is positive is $0 \le x \le 4$.

(v) Finding $a$ and $b$. We are given $y = -x^2 + ax + b$. We can substitute values from the table to find $a$ and $b$.

When $x=0$, $y=1$. Substituting, we have:

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

$b = 1$

So $y = -x^2 + ax + 1$.

Now, when $x=1$, $y=4$. Substituting, we have:

$4 = -(1)^2 + a(1) + 1$

$4 = -1 + a + 1$

$4 = a$

$a = 4$

Therefore, $y = -x^2 + 4x + 1$.

3. Final Answer

(i) $y = 4$ when $x=3$

(ii) Graph sketch is implied by the data and the axis of symmetry.

(iii) (a) Axis of symmetry: $x = 2$ (b) Turning point: $(2, 5)$

(iv) $0 \le x \le 4$

(v) $a = 4$, $b = 1$