Exam practice

$y = (x - 0)^2 + 0$: a = 1, vertex O(0, 0), axis x = 0, opens upward. Points: (±1, 1), (±2, 4).

$y = 2(x - 0)^2 + 0$: vertex O(0, 0), axis x = 0, opens upward, twice as tall as $y = x^2$ (narrower). Points: (±1, 2), (±2, 8).

$y = -3(x - 0)^2 + 0$: vertex O(0, 0), the highest point; axis x = 0; opens downward, three times as tall as $y = x^2$. Points: (±1, −3).

$y = (x - 0)^2 + 2$: vertex (0, 2), axis x = 0, opens upward; $y = x^2$ moved 2 up. It never cuts the X-axis. Points: (±1, 3), (±2, 6).

$y = (x - 0)^2 - 5$: vertex (0, −5), axis x = 0, opens upward; $y = x^2$ moved 5 down. x-intercepts: $x^2 = 5$, $x = \pm\sqrt{5} \approx \pm 2.24$.

$y = (x - 3)^2 + 0$: vertex (3, 0), axis x = 3, opens upward; $y = x^2$ moved 3 right. y-intercept 9; partner (6, 9). Points (1, 4), (5, 4).

$y = \{x - (-3)\}^2 + 0$: vertex (−3, 0), axis x = −3, opens upward; $y = x^2$ moved 3 left. y-intercept 9; partner (−6, 9).

$y = (x + 1)^2 + 0$: vertex (−1, 0), axis x = −1, opens upward; it touches the X-axis at its vertex. y-intercept 1; partner (−2, 1).

This is Example 2: $y = (x - 2)^2 - 1$, vertex (2, −1), axis x = 2, opens upward, cuts the X-axis at (1, 0) and (3, 0), the Y-axis at (0, 3).

Half of 3 is 3/2; take out −1 first.

$$\begin{aligned} y &= -(x^2 + 3x) + 2 \\ &= -\left\{\left(x + \frac{3}{2}\right)^2 - \frac{9}{4}\right\} + 2 \\ &= -\left(x + \frac{3}{2}\right)^2 + \frac{17}{4} \end{aligned}$$
Vertex (−3/2, 17/4) = (−1.5, 4.25), the highest point; axis x = −1.5; opens downward. y-intercept 2, partner (−3, 2). x-intercepts where $(x + 1.5)^2 = 4.25$: $x \approx 0.56$ and −3.56.

$$y = x^2 - x + \frac{1}{4} - \frac{1}{4} - 2 = \left(x - \frac{1}{2}\right)^2 - \frac{9}{4}$$
Vertex (1/2, −9/4) = (0.5, −2.25), axis x = 0.5, opens upward. y-intercept −2, partner (1, −2). x-intercepts: (x − 2)(x + 1) = 0, at (2, 0) and (−1, 0).

$$\begin{aligned} y &= -(x^2 - 7x) + 12 \\ &= -\left\{\left(x - \frac{7}{2}\right)^2 - \frac{49}{4}\right\} + 12 \\ &= -\left(x - \frac{7}{2}\right)^2 + \frac{97}{4} \end{aligned}$$
Vertex (7/2, 97/4) = (3.5, 24.25), the highest point; axis x = 3.5; opens downward. y-intercept 12, partner (7, 12). x-intercepts where $(x - 3.5)^2 = 24.25$: $x \approx 8.42$ and −1.42. (Each grid square below is 2 units.)

• i.$x^2 - 4x + 4 - 4 + 7 = (x - 2)^2 + 3$, so the parabola is $y = (x - 2)^2 + 3$.
• ii.Vertex (2, 3).
• iii.Axis x = 2.
• iv.$y = x^2$ translated 2 units right and 3 units up. Its lowest point is 3 above the X-axis, so the equation $x^2 - 4x + 7 = 0$ has no real solution.

• i.$x^2 - 6x + 13 = (x - 3)^2 - 9 + 13 = (x - 3)^2 + 4$.
• ii.Vertex (3, 4).
• iii.Axis x = 3.
• iv.$y = x^2$ translated 3 units right and 4 units up.

• i.$x^2 - 8x + 21 = (x - 4)^2 - 16 + 21 = (x - 4)^2 + 5$.
• ii.Vertex (4, 5).
• iii.Axis x = 4.
• iv.$y = x^2$ translated 4 units right and 5 units up.

• i.$x^2 + 2x - 4 = (x + 1)^2 - 1 - 4 = (x + 1)^2 - 5$.
• ii.Vertex (−1, −5).
• iii.Axis x = −1.
• iv.$y = x^2$ translated 1 unit left and 5 units down. This one cuts the X-axis, where $(x + 1)^2 = 5$: $x \approx 1.24$ and −3.24.

• i.$x^2 + 2x + 1 = (x + 1)^2 + 0$.
• ii.Vertex (−1, 0).
• iii.Axis x = −1.
• iv.$y = x^2$ translated 1 unit left; it touches the X-axis at its vertex, so the equation has the one solution x = −1.

$x^2 = -2x + 3$: draw $y = x^2$ and the line y = −2x + 3 (through (0, 3), (1, 1), (−3, 9)).

They meet where x = −3 and x = 1. x = 1 or −3. Check: 1 + 2 − 3 = 0 ✓, 9 − 6 − 3 = 0 ✓.

Draw $y = x^2 - 5x + 6$: vertex h = 5/2, k = (24 − 25)/4 = −1/4, so (2.5, −0.25); y-intercept 6. Values: x = 1, 2, 3, 4 give y = 2, 0, 0, 2.

x = 2 or 3.

$x^2 = 4x - 3$: draw $y = x^2$ and y = 4x − 3 (through (0, −3), (1, 1), (3, 9)).

x = 1 or 3, as in Example 2’s x-intercepts.

Draw $y = 2x^2 - 7x + 3$: h = 7/4 = 1.75, $k = (24 - 49)/8 = -25/8 \approx -3.13$. Values: x = 0, 1, 2, 3, 4 give y = 3, −2, −3, 0, 7.

x = 1/2 or 3. Check: 2 × 1/4 − 7/2 + 3 = 0 ✓; 18 − 21 + 3 = 0 ✓.

Draw $y = x^2 - 3x - 10$: h = 3/2, k = (−40 − 9)/4 = −49/4 = −12.25; y-intercept −10. Values: x = −2, 0, 3, 5 give 0, −10, −10, 0.

x = −2 or 5. Check: 4 + 6 − 10 = 0 ✓, 25 − 15 − 10 = 0 ✓.

(−1, 1) and (2, 4). Check: $x^2 = x + 2$ gives (x − 2)(x + 1) = 0 ✓.

x − y = 0 is y = x. The parabola is the one of Example 5, vertex (0.5, −3.25).

(−1, −1) and (3, 3). Check: $x^2 - x - 3 = x$ gives $x^2 - 2x - 3 = 0$, x = −1 or 3 ✓.

Vertex of the parabola: h = 2/12 = 1/6, $k = (-360 - 4)/24 \approx -15.2$. Values: x = −2, −1, 0, 1, 2, 3 give y = 13, −7, −15, −11, 5, 33. The line passes through (0, −3), (−1, −7), (2, 5). (Each grid square is 2 units.)

(−1, −7) and (2, 5). Check: $6x^2 - 2x - 15 = 4x - 3$ gives $6x^2 - 6x - 12 = 0$, $x^2 - x - 2 = 0$, x = 2 or −1 ✓.

The parabola: vertex (1, −1), through O and (2, 0).

(−1, 3) and (4, 8). Check: $x^2 - 2x = x + 4$ gives $x^2 - 3x - 4 = 0$, (x − 4)(x + 1) = 0 ✓.

It cuts the X-axis at −3 and −2: y = a(x + 3)(x + 2). At D(−1, 2): a × 2 × 1 = 2, a = 1.
$y = (x + 3)(x + 2) = x^2 + 5x + 6$. Check with A(−4, 2): 16 − 20 + 6 = 2 ✓.

y = a(x + 4)(x − 3). At D(1, −10): a × 5 × (−2) = −10, a = 1.
$y = (x + 4)(x - 3) = x^2 + x - 12$. Check with C(−2, −10): 4 − 2 − 12 = −10 ✓.

No x-intercepts are given. Use $y = ax^2 + bx + c$ with three of the points; C and D are at the same height, so the axis is halfway between them.

At C(−1, −1): a − b + c = −1 … (i). At D(2, −1): 4a + 2b + c = −1 … (ii). At B(4, 9): 16a + 4b + c = 9 … (iii).
(ii) − (i): 3a + 3b = 0, so b = −a. (iii) − (ii): 12a + 2b = 10, so 10a = 10, a = 1, b = −1. From (i): 1 + 1 + c = −1, c = −3.
$y = x^2 - x - 3$. Check with A(−3, 9): 9 + 3 − 3 = 9 ✓.

D(0, −8) is on the Y-axis: c = −8. At C(−2, −8): 4a − 2b − 8 = −8, so b = 2a. At B(3, 7): 9a + 3b − 8 = 7, so 9a + 6a = 15, a = 1, b = 2.
$y = x^2 + 2x - 8$, the parabola of Example 6. Check with A(−5, 7): 25 − 10 − 8 = 7 ✓.