Transcript Quadratic Formula - chss

Quadratic Formula
Sam Scholten
Graphing Standard Form
Graphing Standard form: Standard form in Quadratic
functions is written as:
Y = ax2+bx+c. The parabola will open up if a>0 and down if
a<0. It will be wider than the graph y = x2 if |a|<1 and
narrower if |a|>1. The x coordinate of the vertex is –b/2a.
The axis of symmetry is the vertical line x = -b/2a.
Y = x2:
Examples
Y = 2x2 – 8x + 6
The coefficients for this function are a = 2, b = -8 and c = 6.
Since a > 0, the parabola opens up. To find the vertex, the
x coordinate is: x = 8/4 or 2. The y coordinate is:
2(2)2 – 8(2) + 6 or -2. The vertex is (2,-2) and the axis of
symmetry is x = 2. Go up 2 and over 1 on each side of
the symmetry so you get the points (1,0) and (3,0). Go
up 6 more and 1 more over on both sides to get (0,6)
and (4,6) (since is 2 up and over 1 the first time, you get
3 times the y increase the second time, but no change
on the x). Draw you parabola through the points.
RESULT
Graphing Vertex Form
Graphing Vertex Form: In vertex form, a quadratic function
is written as:
y = a(x-h)2 + k, where the vertex is (h, k) and the axis of
symmetry is x = h.
Y = -1/2(x + 3)2 + 4:
EXAMPLES
Y = -1/2(x + 3) 2 + 4
a = -1/2, h = -3 and k = 4 and since a < 0, the parabola
opens down.
To graph, plot the vertex (h,k) which is (-3,4) and draw the
axis of symmetry which is x = -3. Since it is -1/2, we will
be going negative two down and two to the side on both
sides. Then you multiply the y axis by three, so six more
down and two more over.
Result
Graphing Intercept Form
Graphing Intercept Form: The intercept form of a quadratic
function is y= a(x-p)(x-q), in this formula, the P and the Q
are x-intercepts and the axis of symmetry is halfway
between (P,0) and (0,Q).
Y = -(x + 2)(x – 4):
EXAMPLES
Y = -(x + 2)(x – 4)
a = -1, p = -2 and q = 4
The x intercepts and p and q so you go halfway between
them to find the axis of symmetry which is x = 1. The x
coordinate of the vertex is one, so we plug that into the
equation and get nine for the y coordinate. Our vertex is
(1,9).
You go down one and over one for the first point on both
sides. Then multiply the y axis by three and you go down
three more and over one more. ( you can also think of it
as adding to the y axis the number two times the first y
coordinate. EX: (1,4) being the first, the next point would
be 2,12 and the next would be (3,20)
RESULT
Solving Quadratics With Square
Roots
Square roots: The product property of square roots is
written as: √ab = √a * √b
The quotient Property is: √a/b = √a/√b provided that in both,
a>0 and b>0.
The square root of a negative number (if r is a real positive
number) is shown as i√r which means (i√r)2 = -r
Square roots can be used in many ways in math. If s > 0,
then x2 = s has two solutions. x = √s and x = -√s which
will often be written like x = +√s which means both.
(i stands for imaginary unit)
EXAMPLE1
3x 2 + 10 = -26
Subtract 10 from both sides (3x 2 = -36)
Divide by 3 (x 2 = -12)
Take the squares of both sides (x = +√-12)
Write in terms of i (x = +i√12)
Simplify the radical (x= +2i√3)
The solutions are -2i√3 and 2i√3
EXAMPLE 2
-3x 2 + 6 = 15
Subtract 6 from both sides ( -3x 2 = 9)
Divide by -3, (x 2 = -3)
Take the squares (x = √-3)
Terms of i, (x = +i√-3)
The solutions are -i√-3 and i√-3
More Complex Numbers
Standard form of a complex number is written as: a + bi
where a is the real number and bi is the imaginary part.
In a complex plane, the horizontal axis is all the real
numbers, and the vertical axis is complex numbers.
These are called the Real axis, and the Imaginary axis.
Plot: 2 – 3i, -3 + 2i, and 4i:
(0,4i)
(-3 + 2i)
(2 – 3i)
Even More Complex Numbers
Complex numbers that are written as: a + bi then a – bi,
are called complex conjugates, which, when multiplied
together, form a real number. You can also use the
conjugates to show the quotient of two complex numbers
in standard form.
The absolute value of a complex number is shown as:
z = a + bi, and |z| is a non-negative real number defined
as: |z| = √(a2 + b2).
Solving Quadratics By Factoring
Solving quadratics with factoring: a2- b2 = (a + b)(a – b) is
the difference of two squares.
A perfect square trinomial is written as:
a2 + 2ab + b2 = (a + b) 2 or: a2 - 2ab + b2 = (a - b) 2
The zero product property is: if A and B are real numbers
or algebraic expressions and AB = 0 then either A or B
will be equal to 0.
In a standard quadratic equation,
x2 + bx + c = (x + m)(x + n) = x + (m + n)x + mn
The sum of n and m must be equal to b and the product
must be equal to c
Example 1
3x2 – 17x + 10
You want 3x2 – 17x + 10 to equal (kx + m) (lx + n) where l
and k are factors of 3 and m and n are (negative) factors
of 10.
There are several options but only (3x – 2)(x – 5) is correct.
Special patterns
1. The difference of 2 squares:
4x2 – 25 = (2x) 2 – 52 equals (2x + 5)(2x – 5)
2. Perfect square trinomial:
9y2 + 24y + 16 = (3y) 2 + 2(3y)(4) + 42 equals (3y + 4)2
Solving Quadratics By Completing
The Square
Completing the square is a process that allows you to write
the form x 2 + bx.
x
x
b
x
x
x2
(b/2)x
(b/2)x
This models the following rule: x2 + bx + (b/2) 2 = (x + b/2) 2
Examples
1. x2 – 7x + c
You get c by taking half of b which in this case is -7
c = (b/2) 2 = ( -7/2) 2 = 49/4
So the original equation is now x2 – 7x + 49/4
So the square is now (x - 7/2) 2
2. x 2 + 10x – 3 = 0, add 3 to both sides
x 2 + 10x = 3, add (10/2) 2 to both sides
x 2 + 10x + 5 2 = 3 + 25, write the left as a binomial squared
(x + 5 ) 2 = 28, take the square roots of both sides
x + 5 = +√28, solve for x
x = -5 + 2√7, the solutions are x = -5 + 2√7 and x = -5 – 2√7
Solving By Using The Quadratic
Formulas
There are 4 types of quadratic inequalities
Y < ax2 + bx + c
Y > ax2 + bx + c
Y > ax2 + bx + c
Y < ax2 + bx + c
Draw the parabola making sure to have a dashed line for
and < and a solid line for > and <
Choose a point inside the parabola and check whether or
not it’s a solution.
If the point is a solution then shade the region inside the
Parabola, if not, shade the outside.
Example 1&2
y > x 2 – 2x – 3
Since its >, the parabola is dashed instead of solid.
Test the point (1,0) in the equation. 0 > 1 – 2 – 3 is 0 > -4
Which is a solution. The graph will be a dotted line with it
shaded inside with a vertex of (1, -4)
y < -2x 2 – 2x – 3
Since its <, the parabola is solid.
Test the point (1, -6) which would make -6 < -2 – 2 – 3 is
-6 < -7 which is not a solution making the graph be a solid
line opening down and wider than a standard graph
with a vertex of (1, -4) and shaded on the outside.
EXAMPLE 3
y > 2x 2 – 2x – 3
Since it is > this will be a dashed line.
Plug in (2,4) which makes: 4 > 8 – 4 – 3 is 4 > 1 which is
true which makes it shaded on the inside of a dashed
graph with a vertex of (1, -4)
Word Problem
You want to plant a rectangular garden along part of a 40
foot side of your house. To keep out animals, you will
enclose the garden with wire mesh along its three open
sides. You will also cover the garden with mulch. You
have 50 ft of mesh and enough mulch to cover 100 sq. ft.
What should the garden’s dimensions be?
Solution
X will be the sides perpendicular to the house and 50 – 2x is
x
the length of the third side.
X(50 – 2x) = 100, Length * width = area
x
50x – 2x 2 = 100,
distributive property
50 – 2x
2
2
-2x + 50x = 100, Write the x -term first
X 2 – 25x = -50,
Divide each side by -2
X 2 – 25x + (-12.5) 2 = -50 + 156.25, Complete the square
(x – 12.5) 2 = 106.25,
Write as binomial squared
X – 12.5 = + √106.25,
Take the square roots of both sides
X = 12.5 + √106.25,
Solve for x
X = 22.8 ft,
Use a calculator to solve
50 – 45.6 = 4.4
So the garden is 22.8 by 4.4 ft.
Discriminant
In the quadratic formula, b 2 – 4ac under the radical sign is
the discriminant which can tell you how many of what
kind of solutions the equation has.
If b 2 – 4ac > 0 the equation has 2 real solutions
If b 2 – 4ac = 0 the equation has 1 real solution
If b 2 – 4ac < 0 the equation has 2 imaginary solutions
The standard form of a quadratic equation’s solution is:
X = (-b + √b 2 – 4ac)/2a