Quadratic Graphs Parabolas 1 To draw a quadratic

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Quadratic Graphs - Parabolas 1. To draw a quadratic graph from y = ax

Quadratic Graphs - Parabolas 1. To draw a quadratic graph from y = ax 2 + bx + c 2. Make up a table of values 3. Draw a smooth curve through the points plotted 4. The shape created is called a parabola

Quadratic Graphs • Note: - X - = + • y = x 2

Quadratic Graphs • Note: - X - = + • y = x 2 x x 2 y -2 4 4 • So -22 = -2 X -2 = 4 -1 1 1 0 0 0 t a h W u o y do k thin e b l l i w the s e u l va 1 1 1 2 4 4 • Use it to find the values of x 2

Quadratic Graphs • y = x 2 x x 2 y -2 4 4

Quadratic Graphs • y = x 2 x x 2 y -2 4 4 -1 1 1 0 0 0 1 1 1 2 4 4 Now join the points with a smooth curve. e h t lot s) of p to nate to d ee rdi rid n We s (coo the g h. p e n a u o r l va nd y the g x a raw d y = x 2 What happens if y = x 2 + 1 or y = x 2 – 1. Let’s Find out.

Quadratic Graphs • y= x x 2 y • -2 4 4 x 2

Quadratic Graphs • y= x x 2 y • -2 4 4 x 2 -1 1 1 0 0 0 1 1 1 2 4 4 y = x 2+1 x -2 -1 0 1 2 x 2 4 1 0 1 4 x 2+1 5 5 2 2 1 1 2 2 5 5 y y = x 2 - 1 x -2 -1 0 1 2 x 2 4 1 0 1 4 x 2 -1 3 0 -1 0 3 y 3 0 -1 0 3 • Note: - x - = + • So (-2)2 = -2 x -2 = 4 Remember x 2 is the same as before. All we need is to add 1 to x 2 to find y. e h t o d n e s ’ h t Le for w e 2 -1 m sa = x y Let’s plot all the three graphs on the grid

Quadratic Graphs • y = x 2 x -2 -1 x 2 4 1

Quadratic Graphs • y = x 2 x -2 -1 x 2 4 1 y 4 1 • y = x 2 + 1 0 0 0 1 1 1 2 4 4 y = x 2 +1 y = x 2 x -2 -1 0 1 2 x 2 4 1 0 1 4 x 2+1 5 2 1 2 5 y 5 2 1 2 5 x -2 -1 0 1 2 x 2 4 1 0 1 4 x 2 -1 3 0 -1 0 3 What do you notice curves? y 3 0 -1 0 3 What do you notice about where the lines cross the y-axis? • y = x 2 - 1

Quadratic Graphs • y = x 2 + 2 x - 2 x -3

Quadratic Graphs • y = x 2 + 2 x - 2 x -3 -2 -1 0 1 x 2 2 x -2 -2 -2 y 2 -2 • Remember that -2 in the equation is the intercept so it will run through

Quadratic Graphs • y = x 2 + 2 x - 2 x -3

Quadratic Graphs • y = x 2 + 2 x - 2 x -3 -2 x 2 9 4 2 x -6 -4 -2 -2 -2 y -1 1 0 0 1 1 -2 -2 0 2 2 4 4 -2 -2 -2 No t So e: - X -2 2 =- - =+ 2 X Use -2 = i val t to 4 ues fin of x 2 d th e Now let’s multiply each value of x by 2 to find the values of 2 x +X-=so 2 X -3 =-6

Quadratic Graphs • y = x 2 + 2 x - 2 x -3

Quadratic Graphs • y = x 2 + 2 x - 2 x -3 x 2 9 2 x -6 -2 -2 y 1 -2 4 -4 -2 -2 -1 1 -2 -2 -3 0 0 0 -2 -2 1 1 2 -2 1 2 4 4 -2 6 To find the values of y, we must sum all the values of x 2, 2 x and -2. Be careful with the negative signs.

Quadratic Graphs • y = x 2 + 2 x - 2 x -3

Quadratic Graphs • y = x 2 + 2 x - 2 x -3 x 2 9 2 x -6 -2 -2 -2 4 -4 -2 -1 1 -2 -2 0 0 0 -2 1 1 2 -2 2 4 4 -2 y -2 -3 -2 1 6 1 Plot the values of x and y on the grid and join the points with smooth curve. y = x 2 + 2 x - 2

Quadratic Graphs • y = x 2 – 3 x - 2 x -1

Quadratic Graphs • y = x 2 – 3 x - 2 x -1 x 2 1 -3 x -2 -2 More negative values 0 1 2 3 4 0 1 4 9 16 -2 -2 -2 • Remember that -2 in the equation is the intercept so it will run through y Note: - X - = + So -12 = -1 X -1 = 1 2. These x f o s e lu a v e Use it to find th mbers. u n e r a u sq y r a in d r o e are just th

Quadratic Graphs • y = x 2 – 3 x - 2 x x

Quadratic Graphs • y = x 2 – 3 x - 2 x x 2 -3 x -2 y -1 1 3 -2 2 0 0 0 1 1 -3 2 4 -6 -2 -2 -2 -4 More negative values 3 4 9 16 -9 -12 -2 -2 -2 2 • Multiply each value of x by -3. -X-=+ -X+=- m su st u m e w , y f o s e lu a v To find the -2. d n a x 2 , 2 x f o s e lu a v e all th ve signs. Be careful with negati.

Quadratic Graphs More negative values • y = x 2 – 3 x -

Quadratic Graphs More negative values • y = x 2 – 3 x - 2 x -1 x 2 1 -3 x 3 0 0 0 1 1 -3 2 4 -6 3 9 -9 4 16 12 -2 y -2 -2 -2 -4 -2 -2 -2 2 Plot the values of x and y on the grid and join the points with smooth curve y = x 2 – 3 x - 2

y = x 2 + 2 x -15 x -6 -5 -4 y 7

y = x 2 + 2 x -15 x -6 -5 -4 y 7 0 -7 -3 -12 -2 -15 -1 -16 0 -15 1 -12 2 3 4 -7 0 7

y = 12 – x - x 2 x y -5 -8 -4 0

y = 12 – x - x 2 x y -5 -8 -4 0 -3 6 -2 10 -1 12 0 12 1 10 2 6 3 4 0 -8

Quadratic Equations All quadratic expressions involve x 2: y = ax 2 + bx

Quadratic Equations All quadratic expressions involve x 2: y = ax 2 + bx + c The graphs of Quadratics are always parabolas x 2 positive x 2 negative The points at which these graphs cut the x-axis are called the solution or roots of the equation

Solving Quadratic Equations from Graphs y = x 2 – 4 x = x(x

Solving Quadratic Equations from Graphs y = x 2 – 4 x = x(x – 4) = 0 x = 0 or x = 4 Roots of the equation

Solving Quadratic Equations from Graphs y = x 2 + 2 x - 3

Solving Quadratic Equations from Graphs y = x 2 + 2 x - 3 = (x + 3)(x – 1) = 0 x = -3 or x = 1 Roots of the equation

Solving Quadratic Equations from Graphs y = 24 - 5 x - x 2

Solving Quadratic Equations from Graphs y = 24 - 5 x - x 2 = (8 + x)(3 – x) = 0 x = -8 or x = 3 Roots of the equation

Solving Quadratic Equations from Graphs y = 5 - 3 x - 2 x

Solving Quadratic Equations from Graphs y = 5 - 3 x - 2 x 2 = (5 + 2 x)(1 - x) Roots of the equation (5 + 2 x)(1 - x) = 0 x = -2· 5 or x = 1

Factorise and Solve ( find the roots of) Mixture of examples

Factorise and Solve ( find the roots of) Mixture of examples

Roots When a number is multiplied by 0 the result is 0. If 7

Roots When a number is multiplied by 0 the result is 0. If 7 b = 0 then b = 0 If ab = 0 then a = 0 or b = 0 If (x - a)(x - b) = 0 then either (x - a) = 0 or (x - b) = 0 x = a or x=b 3 x(x + 4) = 0 3 x = 0 or x + 4 = 0 x = 0 or x = -4 (x – 7)(x + 4) = 0 x – 7 = 0 or x + 4 = 0 x = 7 or x = -4

Mixture of examples Set 1 1. 3 x 2 – 7 x = 0

Mixture of examples Set 1 1. 3 x 2 – 7 x = 0 x(3 x – 7) = 0 2. 4 x 2 – 9 = 0 (2 x + 3)(2 x – 3) = 0 x = 0 or 3 x – 7 = 0 2 x + 3 = 0 or 2 x – 3 = 0 x = 0 or x = 7/3 x = - 3/2 or x = 3/2 3. x 2 - 7 x + 10 = 0 (x - 2)(x – 5) = 0 4. x 2 - 2 x - 48 = 0 (x - 8)(x + 6) = 0 x – 2 = 0 or x – 5 = 0 x – 8 = 0 or x + 6 = 0 x = 2 or x = 5 x = 8 or x = -6

In each example factorise then solve the equation. 1. 5 x 2 – 10

In each example factorise then solve the equation. 1. 5 x 2 – 10 x = 0 5 x(x – 2) = 0 2. x 2 – 16 = 0 x = 0, 2 3. x 2 - 7 x - 18 = 0 (x - 9)(x + 2) = 0 x = -2, 9 5. 100 x 2 – 49 = 0 (10 x - 7)(10 x + 7) = 0 x = -0· 7, 0· 7 7. x 2 + 6 x - 16 = 0 (x - 2)(x + 8) = 0 9. 9 x 2 – 4 = 0 (3 x - 2)(3 x + 2) = 0 x = -8, 2 x = -2/3, 2/3 11. x 2 - 6 x + 8 = 0 (x - 2)(x - 4) = 0 x = 2, 4 (x + 4)(x – 4) = 0 4. 2 x 2 – 3 x = 0 x(2 x – 3) = 0 x = -4, 4 x = 0, 3/2 6. 6 x 2 + 5 x = 0 x(6 x + 5) = 0 x = 0, -5/6 8. x 2 + 5 x + 6 = 0 (x + 2)(x + 3) = 0 10. 8 x 2 – 2 x = 0 2 x(4 x - 1) = 0 x = -3, -2 x = 0, 1/4 12. x 2 - x - 110 = 0 (x - 11)(x + 10) = 0 x = -10, 11

When we cannot factorise or solve quadratic equations graphically we need to use the

When we cannot factorise or solve quadratic equations graphically we need to use the quadratic formula. If ax 2 + bx + c = 0 then

Example : Solve x 2 + 3 x – 3 = 0 ax 2

Example : Solve x 2 + 3 x – 3 = 0 ax 2 + bx + c = 0 1 3 = 0· 79 or – 3· 8 -3

Example 2 Use the quadratic formula to solve the equation : x 2 +

Example 2 Use the quadratic formula to solve the equation : x 2 + 5 x + 6= 0 ax 2 + bx + c= 0 1 5 6 x = - 2 or x = - 3 These are the roots of the equation.

Example 3 Use the quadratic formula to solve the equation : 8 x 2

Example 3 Use the quadratic formula to solve the equation : 8 x 2 + 2 x - 3= 0 ax 2 + bx + c= 0 8 2 -3 x = 1/2 or x = - 3/4 These are the roots of the equation.

Example 4 Use the quadratic formula to solve the equation : 8 x 2

Example 4 Use the quadratic formula to solve the equation : 8 x 2 - 22 x + 15 = 0 ax 2 + bx + c= 0 8 -22 15 x = 3/2 or x = 5/4 These are the roots of the equation.

Example 5 Use the quadratic formula to solve the equation : 2 x 2

Example 5 Use the quadratic formula to solve the equation : 2 x 2 + 3 x - 7 = 0 ax 2 + bx + c= 0 2 3 -7 x = 1· 27 or x = -2· 77 These are the roots of the equation.

Use quadratic formula to solve the following : 2 x 2 + 4 x

Use quadratic formula to solve the following : 2 x 2 + 4 x + 1 = 0 x 2 + 3 x – 2 = 0 x = -1· 7, -0· 3 x = -3· 6, 0· 6 5 x 2 - 9 x + 3 = 0 3 x 2 - 3 x – 5 = 0 x = 1· 4, 0· 4 x = 1· 9, -0· 9 4 x 2 - x - 1 = 0 x 2 + 2 x - 5 = 0 x = -0· 39, 0· 64 x = 1· 45, -3· 45

Consider this rectangle 67 x x+7 The area of the rectangle is given by

Consider this rectangle 67 x x+7 The area of the rectangle is given by A = x(x + 7) So we now obtain the equation x(x+7) = 67 This simplifies to give x 2 + 7 x - 67 = 0 , on moving all the terms to one side of the equation.

 • To solve this equation we use the quadratic equation • So we

• To solve this equation we use the quadratic equation • So we have a = 1 , b = 7 and c = - 67. • This gives

 • This simplifies to • Which leads to • So • Giving the

• This simplifies to • Which leads to • So • Giving the two solutions x = 5· 40 and -12· 40

 • Now x ≠ -12. 40 as you cannot have a negative length

• Now x ≠ -12. 40 as you cannot have a negative length of side so we are left with the solution that x = 5· 40

 • Find x in each of the examples below. 93 x+8 37 x+4

• Find x in each of the examples below. 93 x+8 37 x+4 x+2

Quadratic formula mixed examples The roots of ax 2 + bx + c =

Quadratic formula mixed examples The roots of ax 2 + bx + c = 0 are 1. Solve x 2 - x – 5 = 0 , giving the roots correct to 1 decimal place. 2. Solve 2 x 2 – 5 x + 1, giving the roots correct to 2 decimal places. 3. Solve 2 x(x - 1) = 7 giving the roots correct to 1 decimal place. 4. The area of the rectangle shown is 43 sq. cm. Find x correct to 1 decimal place. x+1 x+5

5. Use Pythagoras to form an equation in x and then solve it. The

5. Use Pythagoras to form an equation in x and then solve it. The triangle is right angled. x x+5 x+3 6. A quadratic equation shown has equation y = x 2 – 8 x + 11. Find the coordinates of the points where the curve cuts the x axis. 7. The height h of the object above the ground after a time t seconds is given by h = 15 t – t 2 When is the object 9 metres above the ground ? Explain the meaning of the two answers.

Quadratics Cutting the x and y axis. In each of the example: a) Find

Quadratics Cutting the x and y axis. In each of the example: a) Find the points where the quadratic cuts the x and the y axis; b) Use symmetry to determine the coordinates of the turning point.

Ex 1. Sketch the graph of y = x 2 + 5 x -

Ex 1. Sketch the graph of y = x 2 + 5 x - 24 Will the graph be “happy” or “sad” x 2 positive, so “happy” Solve x 2 + 5 x – 24 = 0 The graph cuts the x-axis when y = 0 Now factorise (x + 8)(x – 3) = 0 Solving gives x + 8 = 0 or x – 3 = 0 x = -8 or x = 3 Graph cuts y-axis at x = 0 y = 02 +5(0) - 24 Graph cuts the x-axis at (-8, 0) and (3, 0) Cuts y axis at (0, -24) By symmetry the x co-ordinate of turning point is midpoint of the roots -8 and 3. x = (-8 + 3) ÷ 2 = -2· 5 When x = 2· 5, y = 2· 52 + 5(-2· 5) – 24 = 30· 25 Turning point = (-2· 5, -30· 25)

Ex 1. Sketch the graph of y = 15 – 2 x - x

Ex 1. Sketch the graph of y = 15 – 2 x - x 2 Will the graph be “happy” or “sad” x 2 negative, so “sad” Solve 15 – 2 x – x 2 = 0 The graph cuts the x-axis when y = 0 Same as x 2 + 2 x – 15 = 0 Now factorise (x + 5)(x – 3) = 0 Solving gives x + 5 = 0 or x – 3 = 0 x = -5 or x = 3 Graph cuts y-axis at x = 0 Graph cuts the x-axis at y = 15 (-5, 0) and (3, 0) Cuts y axis at (0, 15) By symmetry the x co-ordinate of turning point is midpoint of the roots -5 and 3. x = (-5 + 3) ÷ 2 = -1 When x = -1, y = 15 - 2(-1) – (-1)2 = 16 Turning point = (-1, 16)

Find the points where the quadratic cuts the x and the y axis. Use

Find the points where the quadratic cuts the x and the y axis. Use symmetry to determine the coordinates of the turning point. y 1. y = x 2 – 5 x + 4 (0, 4) C 0 B (4, 0) x A (1, 0) D (2· 5, -2· 25) – 5 x + 4 = 0 (x – 1)(x – 4) = 0 y x 2 x = 1 or x = 4 (1 + 4) ÷ 2 = 2· 5 y = 2· 52 – 5(2· 5) + 4 = 2(2 · 25 · 5, -2· 25) y = x 2 – 3 x - 10 2. A 0 C B D x

3. y y = x 2 + 7 x + 10 A C B

3. y y = x 2 + 7 x + 10 A C B 0 x D y y = x 2 – 8 x + 7 4. D 0 A B x C

y 5. y = x 2 - 9 A 0 B x C y

y 5. y = x 2 - 9 A 0 B x C y 6. y = x 2 – 6 x 0 A B x

7. y y = x 2 – 4 x - 21 0 A B

7. y y = x 2 – 4 x - 21 0 A B C x D y 8. y = x 2 – 6 x + 8 C 0 A D B x

y 9. y = x 2 + 7 x - 8 0 B C

y 9. y = x 2 + 7 x - 8 0 B C A x D 10. y y = x 2 + 4 x + 3 C A D B 0 x

Find the points where the quadratic cuts the x and the y axis. Use

Find the points where the quadratic cuts the x and the y axis. Use symmetry to determine the coordinates of the turning point. 1. y C y = 16 – x 2 y A 0 B x C 2. A 0 D y = 4 + 3 x – x 2 B x

Find the points where the quadratic cuts the x and the y axis. Use

Find the points where the quadratic cuts the x and the y axis. Use symmetry to determine the coordinates of the turning point. 3. D y y = 5 – 4 x – x 2 C A 0 B x y y = -18 + 9 x – x 2 C 4. 0 A B x

Find the points where the quadratic cuts the x and the y axis. Use

Find the points where the quadratic cuts the x and the y axis. Use symmetry to determine the coordinates of the turning point. D y 5. y = -14 – 9 x – x 2 A B 0 C x y = 2 + x – x 2 D y 6. C A 0 B x

Find the points where the quadratic cuts the x and the y axis. Use

Find the points where the quadratic cuts the x and the y axis. Use symmetry to determine the coordinates of the turning point. y 7. C y = 49 – 4 x 2 y A 0 D B 0 8. C A B y = -18 + 11 x – x 2

Find the points where the quadratic cuts the x and the y axis. Use

Find the points where the quadratic cuts the x and the y axis. Use symmetry to determine the coordinates of the turning point. 9. y D y = 10 – 3 x – x 2 C 0 A B B 10. x A y = -5 x – x 2 y B 11. 0 y = 7 x – x 2 y A x 0 x