Find the equation when the vertex is (3,-1) and x-intercepts of 2 and 4
Y= a(x+h) 2 +k
Y=a(x-3) 2 -1
0=a(2-3) 2 -1
0=a(-1) 2 -1
1=a(-1) 2
a=1
So the equation of the line is y= (x-3) 2 -1
Plug the vertex coordinates in for h and k, remember to flip the sign for the h value! Let y=0 and x=2 to solve for a Next we’ll learn some other ways to solve quadratic equations!
4.
Completing the Square (Going from general form to standard form)
Example: Put y=x 2 +10x+23 into standard form
Section off the ‘x’ terms
Y=(x 2 +10x)+23
Add a blank at the end of the bracket
Y=(x 2 +10x+___)+23
Subtract this number from the ‘c’ value
23-25
Y=(x 2 +10x+25)+23-25
Factor the brackets and simplify
Y=(x+5) 2 -2
Hooray! You have now completed the square! 10/2=5 5 2 =25 Plug this number into the blank
Example: If after factoring the solution is (x-3) and (x+2), then the zeroes are 3 and -2. Notice that the zeroes are the x value which will make the solution equal zero
First of all, the equation must be in the form: y=ax 2 +bx+c
Also make sure that the quadratic is in descending powers of x
Note that there will probably be two answers
The Formula: x=-b+- b 2 -4ac 2a Example:2x 2 -5x+2=0 X=-(-5)+- (-5) 2 -4(2)(2) 2(2) X=5+- 9 4 x=2 and x=1/2 These are the zeroes of the equation **For the discriminant remember that no value of x can ever make something in the denominator equal zero
The radicand from the quadratic formula is called the discriminant
b 2 -4ac
We can use the discriminant to determine the nature of the roots
There are three options:
b 2 -4ac>0
b 2 -4ac=0
b 2 -4ac<0
There are two real and distinct roots There are two equal roots There are no real roots **Note: The equation must be in the general form, in descending powers of x
A factor is a value that divides another evenly, the remainder will always be zero
Use the value of the constant to find the factor(s) that will equal zero
Example: Find the factors of x 3 +2x 2 -5x-6 -6 +- 1,2,3,6—Guess and check with calculator, continue until you reach zero F(-1)=0—Now to find the other factors, divide the original equation by (x+1) Since the remainder is zero after the division, there are no more factors
Use this Theorem when the leading coefficient is not 1
x= b/a
b is a factor of the constant
a is a factor of the first term
Example: Find the factors of 6x 3 +13x 2 +x-2 -2 6 +- 1,2 +- 1,2,3,6 F(-1/2)=0 (Therefore (2x+1) is a factor) (2x+1)(6x 2 +10x-4) (2x+1)(3x-1)(x+2) Factor by sum and product:
Is used as a quicker way to divide, easier than using long division
Example: Divide 2x 3 -4x 2 +3x-6 by x+2 2 - Note: Always subtract on the bottom For the top, use the opposite sign of what x equals 2 -4 3 -6 2 -8 19 -44 The factors are: (x+2)(2x 2 -8x+19)-44 (x+2)2(x-2) 2 +11-44 2(x+2)(x-2) 2 -33 x=-2 and x=2 and x=2 The zeroes are:
On your graphing calculator, follow these steps to answer quadratic related questions (in a way that does not involve a pencil!)
Example: Y=-x 2 +3 Y=-x+2 Hit: CALC INTERSECT (Do twice) Scroll the curser to the point of intersection (as close as you can get). Hit guess and record the answer, repeat a second time on the second point of intersection
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