Question Statement

Find the $x$ -intercept, $y$ -intercept, and vertex of the following functions and then plot their graphs:

(i) $f (x) = x^{2} + 2 x + 1$ (ii) $f (x) = - 2 x^{2} + 2 x - 1$ (iii) $f (x) = x^{2} + 2 x$ (iv) $f (x) = 9 - x^{2}$

Background and Explanation

To analyze a quadratic function $f (x) = a x^{2} + b x + c$ , we identify the intercepts and the vertex. The $x$ -intercepts are found by solving $f (x) = 0$ , the $y$ -intercept is found by evaluating $f (0)$ , and the vertex is located at the point where $x = \frac{- b}{2 a}$ .

Solution

(i) $f (x) = x^{2} + 2 x + 1$

For this function, the coefficients are $a = 1, b = 2, c = 1$ .

1. Find the $x$ -intercept: To find where the graph crosses the $x$ -axis, we set $f (x) = 0$ : $x^{2} + 2 x + 1 (x + 1)^{2} x + 1 x = 0 = 0 = 0 = - 1$ The $x$ -intercept is $- 1$ .

2. Find the $y$ -intercept: To find where the graph crosses the $y$ -axis, we set $x = 0$ : $x f (0) y -intercept = 0 = (0)^{2} + 2 (0) + 1 = 1 = 1$

3. Find the Vertex: First, find the axis of symmetry using $x = \frac{- b}{2 a}$ : $x = \frac{- 2}{2 ( 1 )} = - 1$ Now, substitute $x = - 1$ back into the original function to find the $y$ -coordinate: $f (- 1) = (- 1)^{2} + 2 (- 1) + 1 = 1 - 2 + 1 = 0$ The vertex is $(- 1, 0)$ . Since $a > 0$ , the parabola opens upwards.

4. Table of Values and Plot:

$x$	-4	-3	-2	-1	0	1	2
$f (x)$	9	4	1	0	1	4	9

(ii) $f (x) = - 2 x^{2} + 2 x - 1$

For this function, the coefficients are $a = - 2, b = 2, c = - 1$ .

1. Find the $x$ -intercept: Set $f (x) = 0$ : $- 2 x^{2} + 2 x - 1 2 x^{2} - 2 x + 1 = 0 = 0$ Using the quadratic formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ : $x x x = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( 2 ) ( 1 )}{2 ( 2 )} = \frac{2 \pm 4 - 8}{4} = \frac{2 \pm - 4}{4}$ Since we cannot take the square root of a negative number in real numbers, there are no real $x$ -intercepts.

2. Find the $y$ -intercept: Set $x = 0$ : $f (0) = - 1$ The $y$ -intercept is $- 1$ .

3. Find the Vertex: Axis of symmetry: $x = \frac{- b}{2 a} = \frac{- 2}{2 ( - 2 )} = \frac{- 2}{- 4} = \frac{1}{2}$ Find the $y$ -coordinate: $f (\frac{1}{2}) = - 2 (\frac{1}{2})^{2} + 2 (\frac{1}{2}) - 1 = - 2 \times \frac{1}{4} + 1 - 1 = - \frac{1}{2}$ The vertex is $(\frac{1}{2}, - \frac{1}{2})$ .

4. Table of Values and Plot:

$x$	-3	-2	-1	0	1	2	3
$f (x)$	-25	-13	-5	-1	-1	-5	-13

(iii) $f (x) = x^{2} + 2 x$

For this function, $a = 1, b = 2, c = 0$ .

1. Find the $x$ -intercept: Set $f (x) = 0$ : $0 x (x + 2) x = 0 = x^{2} + 2 x = 0 or x = - 2$ The $x$ -intercepts are $(0, 0)$ and $(- 2, 0)$ .

2. Find the $y$ -intercept: Set $x = 0$ : $f (0) = (0)^{2} + 2 (0) = 0$ The $y$ -intercept is $(0, 0)$ .

3. Find the Vertex: Axis of symmetry: $x = \frac{- b}{2 a} = \frac{- 2}{2 ( 1 )} = - 1$ Find the $y$ -coordinate: $f (- 1) = (- 1)^{2} + 2 (- 1) = 1 - 2 = - 1$ The vertex is $(- 1, - 1)$ . Since $a > 0$ , the parabola opens upwards.

4. Table of Values and Plot:

$x$	-3	-2	-1	0	1	2
$f (x)$	3	0	-1	0	3	8

(iv) $f (x) = 9 - x^{2}$

For this function, $a = - 1, b = 0, c = 9$ .

1. Find the $x$ -intercept: Set $f (x) = 0$ : $0 x^{2} (3 - x) (3 + x) x = 3 = 9 - x^{2} = 9 = 0 or x = - 3$ The $x$ -intercepts are $(- 3, 0)$ and $(3, 0)$ .

2. Find the $y$ -intercept: Set $x = 0$ : $f (0) = 9 - (0)^{2} = 9$ The $y$ -intercept is $(0, 9)$ .

3. Find the Vertex: Axis of symmetry: $x = \frac{- b}{2 a} = \frac{- 0}{2 ( - 1 )} = 0$ Find the $y$ -coordinate: $f (0) = 9 - 0 = 9$ The vertex is $(0, 9)$ .

4. Table of Values and Plot:

$x$	-4	-3	-2	-1	0	1	2	3	4
$f (x)$	-7	0	5	8	9	8	5	0	-7

Key Formulas or Methods Used

Standard Form of Quadratic: $f (x) = a x^{2} + b x + c$
$x$ -intercepts: Solve $f (x) = 0$ using factoring or the Quadratic Formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
$y$ -intercept: Evaluate $f (0) = c$
Vertex Formula: $x = - \frac{b}{2 a}$ , then find $y = f (- \frac{b}{2 a})$
Concavity: If $a > 0$ , opens up; if $a < 0$ , opens down.

Summary of Steps

Identify the coefficients $a, b,$ and $c$ from the function.
Set $f (x) = 0$ and solve for $x$ to find the $x$ -intercepts.
Set $x = 0$ to find the $y$ -intercept.
Calculate the $x$ -coordinate of the vertex using $x = - b /2 a$ .
Substitute the $x$ -coordinate back into the function to find the $y$ -coordinate of the vertex.
Create a table of values to determine additional points.
Plot the intercepts and vertex, then draw a smooth curve through the points.

Question Statement

Find the $x$ -intercept, $y$ -intercept, and vertex of the following functions and then plot their graphs:

(i) $f (x) = x^{2} + 2 x + 1$ (ii) $f (x) = - 2 x^{2} + 2 x - 1$ (iii) $f (x) = x^{2} + 2 x$ (iv) $f (x) = 9 - x^{2}$

Background and Explanation

Solution

(i) $f (x) = x^{2} + 2 x + 1$

For this function, the coefficients are $a = 1, b = 2, c = 1$ .

1. Find the $x$ -intercept: To find where the graph crosses the $x$ -axis, we set $f (x) = 0$ : $x^{2} + 2 x + 1 (x + 1)^{2} x + 1 x = 0 = 0 = 0 = - 1$ The $x$ -intercept is $- 1$ .

2. Find the $y$ -intercept: To find where the graph crosses the $y$ -axis, we set $x = 0$ : $x f (0) y -intercept = 0 = (0)^{2} + 2 (0) + 1 = 1 = 1$

4. Table of Values and Plot:

$x$	-4	-3	-2	-1	0	1	2
$f (x)$	9	4	1	0	1	4	9

(ii) $f (x) = - 2 x^{2} + 2 x - 1$

For this function, the coefficients are $a = - 2, b = 2, c = - 1$ .

2. Find the $y$ -intercept: Set $x = 0$ : $f (0) = - 1$ The $y$ -intercept is $- 1$ .

4. Table of Values and Plot:

$x$	-3	-2	-1	0	1	2	3
$f (x)$	-25	-13	-5	-1	-1	-5	-13

(iii) $f (x) = x^{2} + 2 x$

For this function, $a = 1, b = 2, c = 0$ .

1. Find the $x$ -intercept: Set $f (x) = 0$ : $0 x (x + 2) x = 0 = x^{2} + 2 x = 0 or x = - 2$ The $x$ -intercepts are $(0, 0)$ and $(- 2, 0)$ .

2. Find the $y$ -intercept: Set $x = 0$ : $f (0) = (0)^{2} + 2 (0) = 0$ The $y$ -intercept is $(0, 0)$ .

4. Table of Values and Plot:

$x$	-3	-2	-1	0	1	2
$f (x)$	3	0	-1	0	3	8

(iv) $f (x) = 9 - x^{2}$

For this function, $a = - 1, b = 0, c = 9$ .

1. Find the $x$ -intercept: Set $f (x) = 0$ : $0 x^{2} (3 - x) (3 + x) x = 3 = 9 - x^{2} = 9 = 0 or x = - 3$ The $x$ -intercepts are $(- 3, 0)$ and $(3, 0)$ .

2. Find the $y$ -intercept: Set $x = 0$ : $f (0) = 9 - (0)^{2} = 9$ The $y$ -intercept is $(0, 9)$ .

3. Find the Vertex: Axis of symmetry: $x = \frac{- b}{2 a} = \frac{- 0}{2 ( - 1 )} = 0$ Find the $y$ -coordinate: $f (0) = 9 - 0 = 9$ The vertex is $(0, 9)$ .

4. Table of Values and Plot:

$x$	-4	-3	-2	-1	0	1	2	3	4
$f (x)$	-7	0	5	8	9	8	5	0	-7

Key Formulas or Methods Used

Standard Form of Quadratic: $f (x) = a x^{2} + b x + c$
$x$ -intercepts: Solve $f (x) = 0$ using factoring or the Quadratic Formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
$y$ -intercept: Evaluate $f (0) = c$
Vertex Formula: $x = - \frac{b}{2 a}$ , then find $y = f (- \frac{b}{2 a})$
Concavity: If $a > 0$ , opens up; if $a < 0$ , opens down.

Summary of Steps

Identify the coefficients $a, b,$ and $c$ from the function.
Set $f (x) = 0$ and solve for $x$ to find the $x$ -intercepts.
Set $x = 0$ to find the $y$ -intercept.
Calculate the $x$ -coordinate of the vertex using $x = - b /2 a$ .
Substitute the $x$ -coordinate back into the function to find the $y$ -coordinate of the vertex.
Create a table of values to determine additional points.
Plot the intercepts and vertex, then draw a smooth curve through the points.

1.2 Q-3

Question Statement

Background and Explanation

Solution

(i) $f (x) = x^{2} + 2 x + 1$

(ii) $f (x) = - 2 x^{2} + 2 x - 1$

(iii) $f (x) = x^{2} + 2 x$

(iv) $f (x) = 9 - x^{2}$

Key Formulas or Methods Used

Summary of Steps

1.2 Q-3

Question Statement

Background and Explanation

Solution

(i) $f (x) = x^{2} + 2 x + 1$

(ii) $f (x) = - 2 x^{2} + 2 x - 1$

(iii) $f (x) = x^{2} + 2 x$

(iv) $f (x) = 9 - x^{2}$

Key Formulas or Methods Used

Summary of Steps