quadratic functions monika v sikand light and life laboratory department of physics and engineering...

QUADRATIC FUNCTIONS

Monika V Sikand

Light and Life Laboratory

Department of Physics and Engineering physics

Stevens Institute of Technology

Hoboken, New Jersey, 07030.

OUTLINE

Graphing Quadratic Functions

Solving Quadratic Equations by Factoring

Solving Quadratic Equations by Finding Square Roots

Complex Numbers

The Quadratic Formula and the Discriminant

Graphing and Solving Quadratic Inequalities

Modeling with Quadratic Functions

Graphing Quadratic Functions

A QUADRATIC FUNCTION

vertex

Axis of symmetry

y = x2

y = -x2

A quadratic function has a form y = ax2 + bx + c where a ≠ 0.

The graph of a quadratic function is U-shaped and is called parabola.

The lowest or highest point on the graph of a quadratic function is called the vertex.

The graphs of y = x2 and y = -x2 are symmetric about the y-axis, called the axis of symmetry.

THE GRAPH OF A QUADRATIC FUNCTION

vertex

Axis of symmetry

y = x2

y = -x2

The parabola opens up if a>0 and opens down if a<0.

The parabola is wider than the graph of y = x2 if |a| < 1 and narrower than the graph of y = x2 if |a| > 1.

The x-coordinate of the vertex is -b/2a.

The axis of symmetry is the vertical line x = -b/2a.

EXAMPLE

Graph y = 2x2 -8x +6

Solution: The coefficients for this function area = 2, b = -8, c = 6.

Since a>0, the parabola opens up.The x-coordinate is: x = -b/2a, x = -(-8)/2(2) x = 2

The y-coordinate is: y = 2(2)2-8(2)+6 y = -2

Hence, the vertex is (2,-2).

EXAMPLE(contd.)

Draw the vertex (2,-2) on graph.

Draw the axis of symmetry x=-b/2a.

Draw two points on one side of the axis of symmetry such as (1,0) and (0,6).

Use symmetry to plot two more points such as (3,0), (4,6).

Draw parabola through the plotted points.

(2,-2)

(1,0)

(0,6)

(3,0)

(4,6)

Axis of symmetry

x

y

VERTEX FORM OF QUADRATIC EQUATION

y = a(x - h)2 + k

The vertex is (h,k).

The axis of symmetry is x = h.

GRAPHING A QUADRATIC FUNCTION IN VERTEX FORM

(-3,4)

(-7,-4)

(-1,2)(-5,2)

(1,-4)

Axis of symmetry

x

y

Example y = -1/2(x + 3)2 + 4where a = -1/2, h = -3, k = 4. Since a<0 the parabola opens down.

To graph a function, first plot the vertex (h,k) = (-3,4).

Draw the axis of symmetry x = -3

Plot two points on one side of it, such as (-1,2) and (1,-4).

Use the symmetry to complete the graph.

INTERCEPT FORM OF QUADRATIC EQUATION

y = a(x - p)(x - q)

The x intercepts are p and q.

The axis of symmetry is halfway between (p,0) and (q,0).

GRAPHING A QUADRATIC FUNCTION IN INTERCEPT FORM

(-2,0)

(1,9)

(4,0)

Axis of symmetry

x

y

Example y = -(x + 2)(x - 4).where a = -1, p = -2, q = 4. Since a<0 the parabola opens down.

To graph a function, the x - intercepts occur at (-2,0) and (4,0).

Draw the axis of symmetry that lies halfway between these points at x = 1.

So, the x - coordinate of the vertex is x = 1 and the y - coordinate of the vertex is: y = -(1 + 2)(1 - 4)= 9.

WRITING THE QUADRATIC EQUATION IN STANDARD FORM

(1). y = -(x + 4)(x - 9) = -(x2 - 9x + 4x - 36) = -(x2 - 5x -36) = -x2 + 5x + 36

(2). y = 3(x -1)2 + 8 = 3(x -1)(x - 1) + 8 = 3(x2 - x - x + 1) + 8 = 3(x2 - 2x + 1) + 8 = 3x2 - 6x + 3 + 8 = 3x2 - 6x + 11

QUADRATIC FUNCTIONS IN REAL LIFE

Researchers conducted an experiment to determine temperatures at which people feel comfortable. The percent of test subjects who felt comfortable at temperature x( in degrees Fahrenheit) can be modeled by:

y = -3.678x2 + 527.3x - 18.807

a. What temperature made the greatest percent of test subjects comfortable?

b. At that temperature , what percent felt comfortable?

SOLUTION

Since a = -3.678 is negative, the graph of the quadratic function open down and the function has a maximum value. The maximum value occurs at:

The corresponding value of y is:

a. Hence, The temperature that made the greatest percent of test subjects comfortable was about 72.

b. At that temperature about 92% of the subjects felt comfortable.

€

x = −b

2a= −

527.3

2(−3.678)≈ 72

€

y = −3.678(72)2 + 527.3(72) −18,807 ≈ 92

REAL LIFE EXAMPLE 6(page 252)

The Golden Gate Bridge in San Francisco has two towers that rise 500 feet above the road and are connected by suspension cables as shown. Each cable forms a parabola with equation

where x and y are measured in feet.

a. What is the distance d between the towers?b. What is the height l above the road of a cable at its lowest

point?

€

y =1

8940(x − 2100)2 + 8

GOLDEN GATE BRIDGE

l

d

y

200ft

500ft

x

Not down to scale

SOLUTION

€

y =1

8960(x − 2100)2 + 8

y =1

8960(x2 − 4200x+ 4410000)+ 8

y =1

8960x2 −

4200

8960x+(

4410000

8960+ 8)

x =−b

2a= (

4200

8960)(

1

2 ×1

8960

)

x = 2100€

y =1

8960(x − 2100)2 + 8

y =1

8960(2100 − 2100)2 + 8

y =0

8960+ 8

y = 8

Hence the vertex of the parabola is (2100,8)

SOLUTION(contd.)

a. The vertex of the parabola is (2100,8), so the cable’s lowest point is 2100 feet from the left tower shown above. Since the heights of the two tower’s are the same, the symmetry of the parabola implies that the vertex is also 2100 feet from the right tower. Therefore the towers are d = 2(2100) = 4200 feet apart.

b. The height l above the road of a cable at its lowest point is the y-coordinate of the vertex. Since the vertex is (2100,8), this height is l= 8 feet.

SOLVING QUADRATIC EQUATION BY FACTORING

FACTORING QUADRATIC EXPRESSION

The expression x2 + bx + c is a trinomial because it has three terms. We can use factoring to write it as product of two terms or binomials such as

x2 + bx + c = (x+m)(x+n) = x2 + (m+n)x + mn

Example:x2 + 8x + 15 = (x + 3)(x + 5)

EXAMPLE

Problem: Factor x2 - 12x - 28Solution: x2 - 12x - 28 = (x+m)(x+n) where mn = -28 and m+n = -12

Factors of -28

-1,28 1,-28 -2,14 2,-14 -4,7 4,-7

Sum of factors

27 -27 12 -12 3 -3

The table shows that m = 2 and n = -14.So, x2 - 12x - 28 = (x + 2)(x - 14)

SPECIAL FACTORING PATTERNS

1. Difference of two squares: a2 - b2 = (a+b)(a-b)

Example: x2 - 9 = (x+3)(x-3)

1. Perfect square Trinomial: a2 + 2ab + b2 = (a+b)2

Example: x2 + 12x +36 = (x+6)2

3. Perfect square Trinomial: a2 - 2ab + b2 = (a-b)2

Example: x2 -8x +16 = (x-4)2

FACTORING MONOMIALS FIRST

Monomial is an expression that has only one term.

Factor the quadratic expression:

a. 5x2 - 20 = 5(x2 -4) = 5(x+2)(x-2)

• 6p2 + 15p + 9 = 3(2p2 + 5p + 3) = 3(2p + 3)(p + 1)

SOLVING QUADRATIC EQUATIONS

€

x2 + 3x −18 = 0

(x+ 6)(x − 3) = 0

x+ 6 = 0

x = −6

or

x − 3 = 0,

x = 3

Solve: Solve:

€

2t 2 −17t + 45 = 3t − 5

2t 2 − 20t + 50 = 0

t 2 −10t + 25 = 0

(t − 5)2 = 0

t − 5 = 0

t = 5

The solution is 5.The solutions are -6 and 3.

REAL LIFE EXAMPLE

You have made a rectangular stained glass window that is 2 feet by 4 feet. You have 7 square feet of clear glass to create a border of uniform width around the window. What should the width of the border be?

x xx x

xx x

x2

4 4+2x

2+2x

SOLUTION

Let the width of the border be = x Area of the border = 7 Area of the border and window = (2+2x)(4+2x) Area of the window = 24 = 8

Area of border = Area of border & window - Area of window

7 = (2+2x)(4+2x) - 80 = 4x2 + 12x -70 = (2+7x)(2x-1)2+7x = 0 or 2x-1 = 0 x = -3.5 o x = 0.5Rejecting negative value, -3.5. Hence the border’s width is 0.5ft

ZERO PRODUCT PROPERTY

Let A and B be real numbers or algebraic expressions. If AB=0, then either A = 0 or B = 0

Solve (a) x2 + 3x -18 = 0

(x + 6)(x - 3) = 0. Hence either x + 6 = 0 or x - 3 = 0.

The solutions are x = -6 or x = 3

ZERO PRODUCT PROPERTY

Let A and B be real numbers or algebraic expressions. If AB=0, then either A = 0 or B = 0 Solve (a) 2t2 - 17t + 45 = 3t - 5 2t2 - 20t + 50 = 0 t2 - 10t + 25 = 0 (t - 5)2 = 0 t - 5 = 0 t = 5 Hence the solutions is 5.

FINDING ZEROES OF QUADRATIC FUNCTIONS

Find zeros of y = x2 -x -6

Solution: y = x2 - 3x + 2x - (32) y = x(x-3) + 2(x-3) y = (x + 2)(x - 3)

Hence the zeros of the function are -2 and 3.

SOLVING QUADRATIC EQUATIONS BY FINDING

SQUARE ROOTS

SQUARE ROOT

A number r is a square root of a number s if r2 = s. A positive number s has square roots denoted by

Example :Since 32 = 9 and (-3)2 = 9. The square roots of 9 are:

€

sor −s

€

9 = 3

− 9 = −3.

PROPERTIES OF SQUARE ROOTS (a>0, b>0)

PRODUCT PROPERTY:

QUOTIENT PROPERTY:

€

ab = a. b

€

a

b=a

b

SIMPLIFY

€

(a). 24

= 4. 6

= 2 6

€

(b).7

2

=7

2•

2

2

=14

2

Solving a quadratic equation

€

2x2 +1 =17

2x2 =16

x2 = 8

x = ± 8

x = ±2 2

Hence the solutions are:

€

2 2 and

€

−2 2€

1

3(x+ 5)2 = 7

(x+ 5)2 = 21

(x+ 5) = ± 21

x = −5 ± 21

€

−5 + 21

€

−5 − 21

Hence the solutions are:

and

Solve: Solve:

REAL LIFE EXAMPLE

A stunt man working on the set of a movie is to fall out of a window 100 feet above the ground. For the stunt man’s safety, an air cushion 26 feet wide by 30 feet long by 9 feet high is positioned on the ground below the window.

a. For how many seconds will the stunt man fall before he reaches the cushion?

b. A movie camera operating at a speed of 24 frames per second records the stunt man’s fall. How many frames of film show the stunt man falling?

SOLUTION

a. The stunt man’s initial height is ho = 100 feet, so his height as a function of time can be modeled by function h = -16t2 + 100. Since the height of the cushion is 9 feet above the ground, the the time taken by the stunt man to reach the cushion is: h = -16t2 + 100 9 = -16t2 + 100 -91 = -16t2

91/16 = t2

or t ≈ 2.4. Thus, it takes about 2.4 seconds for the stunt man to reach the cushion.

b. The number of frames of film that show the stunt man falling is given by the product (2.4sec)(24 frames/sec), or about 57 frames.

COMPLEX NUMBERS

COMPLEX NUMBER

A complex number written in standard form is a number a+bi where a and b are real numbers.

The number a is the real part of the complex number and number bi is the imaginary part.

If b≠0, then a+bi is an imaginary number. If a=0 and b≠0, then a+bi is a pure imaginary number.

A complex plane has a horizontal axis called the real axis and a vertical axis called the imaginary axis.

THE SQUARE ROOT OF A NEGATIVE NUMBER

1. If r is a positive real number, then

where i is the imaginary unit defined as

2. By property (1), it follows that€

−r = i r

€

(i r )2 = −r€

i = −1

SOLVING A QUADRATIC EQUATION

€

3x2 +10 = −26

Solution

3x2 +10 = −26

3x2 = −36

x2 = −12

x = ± −12

x = ±i 12

x = ±2i 3

Hence, the solutions are and

€

−2i 3

€

2i 3

Solve:

PLOTTING COMPLEX NUMBERS

Plot 2-3i in the complex plane.

To plot 2-3i , start at the origin, move 2 units to the right and then move 3 units down.

0

2-3i

x

y

ADDING AND SUBTRACTING COMPLEX NUMBERS

Sum of complex numbers:

(a+bi) + (c+di) = (a+c) + i(b+d)

Example:(4-i) + (3+2i) = (4+3) + i(-1+2) = 7 + i

Difference of complex numbers:

(a+bi) - (c+di) = (a-c) + i(b-d)

Example:(7-5i) - (1-5i) = (7-1) + i(-5+5) = 6 + 0i = 6

MULTIPLYING THE COMPLEX NUMBERS

Write the expression as a complex number in standard form.

a. 5i(-2+i) = -10i + 5i2 = -10i + 5(-1) = -5-10i

b. (7-4i)(-1+2i) = -7 + 14i + 4i - 8i2

= -7 + 18i - 8(-1) = 1 + 18i

DIVIDING COMPLEX NUMBERS

Write the quotient in standard form.

Solution:Multiply the numerator and denominator by the complex conjugate of the denominator.

€

5 + 3i

1− 2i

€

5 + 3i

1− 2i•

1+ 2i

1+ 2i=

5 +10i+ 3i+ 6i2

1+ 2i − 2i − 4i2

−1+13i5

= −15

+135i

ABSOLUTE VALUES OF COMPLEX NUMBER

Find the absolute value of each complex number.

€

(a)3+ 4i = 32 + 42 = 25 = 5

(b)−2i = 0 +(−2i) = 02 +(−2)2 = 2

(c)−1+ 5i = (−1)2 + 52 = 26 ≈ 5.10

THE QUADRATIC FORMULA AND THE DISCRIMINANT

THE QUADRATIC FORMULA

Let a, b, and c be real numbers such that a≠0. The solutions of the quadratic equation ax2 + bx +c = 0 are:

€

x =−b± b2 − 4ac

2a

SOLVING QUADRATIC EQUATION WITH TWO REAL SOLUTIONS

€

Solve

2x2 + x = 5

Solution :

2x2 + x − 5 = 0

x =−b± b2 − 4ac

2a

x =−1± 12 − 4(2)(−5)

2(2)

x =−1± 41

4

€

x =−1+ 41

4≈ 1.35

and

x =−1− 41

4≈ −1.85

The solutions are:

SOLVING QUADRATIC EQUATION WITH ONE REAL SOLUTIONS

€

Solve

x2 − x = 5x − 9

Solution :

x2 − x = 5x − 9

x2 − 6x+ 9 = 0

x =6 ± (−6)2 − 4(1)(9)

2(1)

x =6 ± 0

2x = 3

Hence, the solution is 3.

SOLVING QUADRATIC EQUATION WITH TWO IMAGINARY SOLUTIONS

€

Solve

−x2 + 2x = 2

Solution

−x2 + 2x = 2

−x2 + 2x − 2 = 0

x =−2 ± 22 − 4(−1)(−2)

2(−1)

x =−2 ± −4

−2

x =−2 ± 2i

−2x =1± i

The solutions are:1+i and 1-i

DISCRIMINANT

In the quadratic formula, the expression b2-4ac under the radical sign is called the discriminant of the associated equation ax2 + bx + c = 0.

€

x =−b± b2 − 4ac

2a

NUMBER AND TYPE OF SOLUTIONS OF A QUADRATIC EQUATION

Consider the quadratic equation ax2 + bx + c = 0.

If b2-4ac > 0, then the equation has two real solutions.

If b2-4ac = 0, then the equation has one real solutions.

If b2-4ac < 0, then the equation has two imaginary solutions.

EXAMPLE: TWO REAL SOLUTIONS

€

Solve

x2 − 6x+ 8 = 0

Discriminant :

b2 − 4ac = (−6)2 − 4(1)(8) = 4

Solutions :

x =−b± b2 − 4ac

2a

x =6 ± (−6)2 − 4(1)(8)

2

x =6 ± 4

2=

6 ± 2

2= 3±1 = 4,2

Hence there are two real solutions: 4,2

EXAMPLE: ONE REAL SOLUTION

€

Solve

x2 − 6x+ 9 = 0

Discriminant :

b2 − 4ac = (−6)2 − 4(1)(9) = 0

Solutions :

x =−b± b2 − 4ac

2a

x =6 ± (−6)2 − 4(1)(9)

2

x =6 ± 0

2=

6

2= 3

Hence, there is one real solution: 3

EXAMPLE: TWO IMAGINARY SOLUTIONS

Find the discriminant of the quadratic equation and give the number

and type of solutions of the equation.

€

Solve

x2 − 6x+10 = 0

Discriminant :

b2 − 4ac = (−6)2 − 4(1)(10) = −4

Solutions :

x =−b± b2 − 4ac

2a

x =6 ± (−6)2 − 4(1)(10)

2

x =6 ± −4

2=

6 ± 2 −1

2= 3± i

Hence there are two imaginary solutions:

3+i and 3-i

REAL LIFE EXAMPLE

A baton twirler tosses a baton into the air. The baton leaves the twirler’s hand 6 feet above the ground and has an initial vertical velocity of 45 feet per second. The twirler catches the baton when it falls back to a height of 5 feet. For how long is the baton in the air?

Solution:Since the baton is thrown, we use the model h = -16t2 + vot + ho where vo = 45, ho = 6, h = 5.

SOLUTION(contd.)

€

h = −16t 2 + vot +ho

5 = −16t 2 + 45t + 6

0 = −16t 2 + 45t +1

t =−45 ± 2089

−32t = −0.022

or

t = 2.8

Rejecting the negative solution, the baton is in the air for about 2.8 seconds.

GRAPHING AND SOLVING QUADRATIC INEQUALITIES

GRAPHING A QUADRATIC INEQUALITY IN TWO VARIABLES

y < ax2 +bx +c y ≤ ax2 +bx +c y > ax2 +bx +c y ≥ ax2 +bx +c

Draw parabola with equation y = ax2 +bx +c . Make the parabola dashed for inequalities with < or > and solid for inequalities with ≤ or ≥ . Choose a point (x,y) inside the parabola and check whether the point is a solution of the inequality. If a point (x,y) is a solution, shade the region inside the parabola. If it is not the solution, shade the region outside the parabola.

EXAMPLE

0

€

•(1,0) x

y

4

1

Graph y > x2 -2x -3

Solution:Graph y = x2 -2x -3. Since the inequality is >, make parabola dashed.Test the point (1,0) inside the parabola. y > x2 -2x -30 > 12-2(1)-30 > -4So, (1,0) is a solution of the inequality.Shade the region inside the parabola.

REAL LIFE EXAMPLE

You are building a wooden bookcase. You want to choose a thickness d(in inches) for the shelves so that each is strong enough to support 60 pounds of books without breaking. A shelf can safely support a weight of W ( in pounds) provided that

W ≤ 300d2

a. Graph the given inequalityb. If you make each shelf 0.75 inch thick, can it

support a weight of 60 pounds?

SOLUTION

y

0

50100150200250300

W

d0.5 1.0 1.5

a. Graph W = 300d2 for non negative values of d. Since the inequality symbol is ≤ make the parabola solid. Test a point inside the parabola, such as (0.5, 240). W ≤ 300d2

240 ≤ 300(0.5)2

240 ≤ 75Since the chosen point is not a solution, shade the region outside (below) the parabola.

b. The point (0.75,60) lies in the shaded region of the graph from part (a), so (0.75,60) is a solution of the given inequality.

c. Therefore, a shelf that is 0.75 inch thick can support a weight of 60 pounds.

GRAPHING A SYSTEM OF QUADRATIC INEQUALITY

x

yGraph the system of quadratic inequalities.y≥x2-4 Inequality 1y<-x2-x+2 Inequality 2

Solution:1. Graph the inequality y≥x2-4. The graph is

in red region inside and including the parabola y = x2-4.

2. Graph the inequality y<-x2-x+2. The graph is in blue region inside (but not including) the parabola y = -x2-x+2.

3. Identify the region where two graphs overlap. This region is the graph of the system.

y≥x2-4

y<-x2-x+2.

QUADRATIC INEQUALITY IN ONE VARIABLE

1. To solve ax2 + bx + c < 0 (or ax2 + bx + c ≤ 0), graph y = ax2 + bx + c and identify the x values for which the graph lies below (or on and below) the x-axis.

2. To solve ax2 + bx + c > 0 (or ax2 + bx + c ≥ 0), graph y = ax2 + bx + c and identify the x values for which the graph lies above (or on and below) the x-axis.

EXAMPLE

x

y

3

1

1 5

Solve x2 - 6x + 5 < 0Solution:1. The solution consist of the x-values for

which the graph of y = x2 - 6x + 5 lies below the x-axis.

2. Find the graph’s x-intercepts by letting y = 0 0 = x2 - 6x + 5 0 = (x-1)(x-5) x = 1 or x = 5Sketch a parabola that opens up and has 1 and 5 as x-intercepts.

3. The graph lies below the x-axis between x = 1 and x = 5.

4. The solution of the given inequality is 1<x<5.

EXAMPLE

Solve 2x2 + 3x -3 ≥ 0Solution:1. The solution consist of the x-values for which the graph

of y = 2x2 + 3x -3 lies on and above the x-axis.2. Find the graph’s x-intercepts by letting y = 0

0 = 2x2 + 3x -3

€

x =−3± 32 − 4(2)(−3)

2(2)

x =−3± 33

4x ≈ 0.69or

x ≈ −2.19

EXAMPLE(contd.)

-2.19 4 x

y

0.69

1

1. Sketch a parabola that opens up and has 0.69 and -2.19 as x-intercepts.

2. The graph lies on and above the x-axis to the left of ( and including) x = -2.19 and to the right of ( and including) x = 0.69.

3. The solution of the given inequality is approximately x ≤ -2.19 or x ≥ 0.69.

SOLVING A QUADRATIC INEQUALITY ALGEBRAICALLY

Solve:x2 + 2x ≤ 8

Solution:First replace the inequality symbol with equal sign.x2 + 2x = 8x2 + 2x - 8 = 0(x+4)(x-2) = 0x = - 4 or x = 2

The numbers -4 and 2 are the critical x-values of the inequality x2 + 2x

≤ 8. Plot -4 and 2 on a number line.

SOLUTION (contd.)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

These critical x-values partition the number line into three intervals.

Test x = -5, (-5)2 + 2(-5) = 15 ≤ 8Test x = 0, (0)2 +2(0) = 0 ≤ 8Test x = -3, (3)2 + 2(3) = 15 ≤ 8Hence the solution is -4 ≤ x ≤ 2.

REAL LIFE EXAMPLE

For a driver aged x years, a study found that the driver’s reaction time V(x) (in milliseconds) to a visual stimulus such as traffic can be modeled by: V(x) = 0.005x2 - 0.23x + 22, 16 ≤ x ≤ 70At what ages does a driver’s reaction time tend to be greater than 25 milliseconds?

Solution:The values of x for which V(x) > 250.005x2 - 0.23x +22 > 250.005x2 - 0.23x - 3 > 0The solution consists of the x- values for which the graph lies above the x-axis.

SOLUTION(contd.)

The graph’s x intercept is found by letting y =0 and using the quadratic formula to solve for x. 0.005x2 - 0.23x - 3 = 0

€

x =0.23± (0.23)2 − 4 × 0.005 ×(−3)

2 × 0.005

x =0.23± .0529 +.06

.01

x =0.23± .1129

.01

x =0.23± 0.34

.01x = 57approx.

or

x = −11approx

Rejecting the negative value, the graph’s x-intercept is about 57. The graph of 0.005x2 - 0.23x - 3 = 0 lies in the domain 16 ≤ x ≤ 70. The graph lies above the x-axis when 57 < x ≤ 70.

Hence the drivers over 57 years old tend to have reaction times greater than 25 milliseconds.

MODELING WITH QUADRATIC FUNCTIONS

QUADRATIC FUNCTION IN VERTEX FORM

(2,-3)

(4,1)

x

y

1

1

Write the quadratic function for the parabola shown.

Solution:The vertex shown is (h,k) = (2,-3)Using the vertex form of the quadratic function.y = a(x-h)2 + ky = a(x-2)2 - 3Use the other given point (4,1) to find a.1 = a(4-2)2 - 31 = 4a - 34 = 4a1 = aHence the quadratic function for the parabola is y = (x-2)2 -3

QUADRATIC FUNCTION IN INTERCEPT FORM

(-1,2)

(-2)

(3)

x

y

1

1

Write the quadratic function for the parabola shown.

Solution:The x intercepts shown are p = -2, q = 3Using the intercept form of the quadratic function.y = a(x-p)(x-q)y = a(x+2)(x-3)Use the other given point (-1,2) to find a.2 = a(-1+2)(-1-3)2 = -4a-1/2 = aHence the quadratic function for the parabola is y = -1/2(x+2)(x-3)

END