The problem provides an incomplete table of values for the function $y = 4 - (x-1)^2$. Several tasks need to be performed: (i) Find the value of $y$ when $x = 1$. (ii) Draw the graph of the function using a suitable scale. (iii) Describe the behavior of the function within the interval $2 < x < 4$. (iv) Express the function in the form $y = (a+x)(b-x)$ and find the values of $a$ and $b$. (v) Write down the solutions of the equation $x^2 = 2x + 3$ using the graph drawn.

1. Problem Description

The problem provides an incomplete table of values for the function $y = 4 - (x-1)^2$. Several tasks need to be performed:

(i) Find the value of $y$ when $x = 1$.

(ii) Draw the graph of the function using a suitable scale.

(iii) Describe the behavior of the function within the interval $2 < x < 4$.

(iv) Express the function in the form $y = (a+x)(b-x)$ and find the values of $a$ and $b$.

(v) Write down the solutions of the equation $x^2 = 2x + 3$ using the graph drawn.

2. Solution Steps

(i) To find the value of $y$ when $x=1$, substitute $x=1$ into the equation $y = 4 - (x-1)^2$:

$y = 4 - (1-1)^2 = 4 - (0)^2 = 4 - 0 = 4$.

(ii) To draw the graph, we need the complete table of values. The given table is:

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

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

y | -5 | 0 | 3 | | 3 | 0 | -5

We found that when $x=1$, $y=4$. So the completed table is:

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

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

y | -5 | 0 | 3 | 4 | 3 | 0 | -5

Plot these points on a graph with the x-axis and y-axis. The appropriate scale can be chosen to accommodate these values. Connect the points with a smooth curve.

(iii) In the interval $2 < x < 4$, the $y$ values are decreasing from 3 to -

5. Thus, the function is decreasing within this interval.

(iv) Expand the given function $y = 4 - (x-1)^2$.

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

We want to express this in the form $y = (a+x)(b-x)$. Expanding this gives

$y = ab - ax + bx - x^2 = -x^2 + (b-a)x + ab$.

Comparing the coefficients, we have $b-a = 2$ and $ab = 3$.

We are given that $y = (a+x)(b-x)$. We also know that when $y = -x^2 + 2x + 3$, we can factor it to $y = (3+x)(1-x) = (x+3)(1-x)$. Thus, we can say $a=3$ and $b=1$.

(v) We are given the equation $x^2 = 2x + 3$. We can rewrite this as $x^2 - 2x - 3 = 0$.

We also know that $y = -x^2 + 2x + 3$, so $-y = x^2 - 2x - 3$.

Thus, we want to find when $y=0$ in our equation. The graph intersects the x-axis when $y=0$. From the completed table, we see that $y = 0$ when $x = -1$ and $x = 3$.

So, the solutions to the equation $x^2 = 2x + 3$ are $x = -1$ and $x = 3$.

3. Final Answer

(i) $y = 4$

(ii) Graph of $y=4-(x-1)^2$ with points (-2, -5), (-1, 0), (0, 3), (1, 4), (2, 3), (3, 0), (4, -5)

(iii) The function is decreasing.

(iv) $a = 3$, $b = 1$

(v) $x = -1$ and $x = 3$