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itutor

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6.6 analyzing graphs of quadratic functions
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### Analytic Geometry Period 1

• 1. Analytic Geometry Opt me In entering the Database of... Analytic Geometry® is a PowerPoint Presentation by The Fantastic Four©. All rights Reserved XD
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• 5. Distance Equations Finding distance can be tricky, but this section will show you just how to solve them.
• 6. Distance Equations HOW TO SOLVE Example: Line1 0 = 4x - 2y - 8 y = 2x - 4 Line2 4y = 8x + 8 y = 2x + 2 Find the horizontal distance for Line1. Go to ‘x’ axis if you can determine the distance between the two intercepts. 3 6 Line2 Line1 Find the vertical distance for Line2. Go to ‘y’ intercepts and find the difference between them. (1, 4) (1, -2) The shortest distance distance formula |Ax + By + C| A + B 2 2 * Continued on next page
• 7. Distance Equations Continued .. When you use the formula, use the ‘x’ and ‘y’ of the perpendicular line. Line1 0 = 4x - 2y - 8 A = 4 B = -2 C = -8 L = |4(1) + -2(4) + -8| 4 + -2 1 2 2 = |4 – 8 – 8| 16 + 4 2 2 = |-12| 20 20 20 = 12 20 20 Line2 0 = 2x - y + 2 A = 2 B = -1 C = 2 L = |2(1) + -1(-2) +2| 2 + -1 2 2 2 = |2 + 2 + 2| 5 = |6| 5 5 5 = 6 5 5
• 8. Two Variable Systems In this section, we will show you how to solve two variable systems.
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• 10. Two Variable Equations Continued .. Step 3: Either add or subtract to eliminate 3x - 2y = 6 - x - 2y = 4 2x 0 = 2 Algebraically solve for ‘x’ 2x = 2 x = 1 Step 4: Substitute the value back into the original equation to solve for second variable 3x - 2y = 6 3(1) - 2y = 6 3 - 2y = 6 -2y = 3 y = -3/2 * Solution is (1, -3/2)
• 11. Three Variable Systems A repeat of two variable systems, but an extra equation is added. A different solution is presented.
• 12. 3 Variable Systems … Oh, three variables, three equations… LOVELY! HOW TO SOLVE; 1) -9x + 5y – 8z = -61 2) -4x + y – 2z = -7 3) 7x – 5y + 11z = 96 Now, to solve this, we must make two more equations ; so using equations 1&2, we can make equation four like so ; -9x + 5y – 8z = -61 5(-4x + y – 2z = -7)  11x + 2z = -26 <- multiplied the equation by 5 so that at least one of the variables have the same value [in this case ‘’y’’] so that we can remove one of the variables to find one of the remaining two. Using equations 1&3, we can make equation 5 ; **Note ; When making equations 4 & 5 be sure that you are eliminating the same variable. -9x + 5y – 8z = -61 +7x – 5y + 11z = 96  -2x + 3z = 35 RULE ; 3 Variables, 3 Equations 4) 11x + 2z = -26 5) -2x + 3z = 35 Now that we have these two, we can now substitute.  3(11x + 2z = -26) 33x + 6z =-78 37x = - 148 2(-2x + 3z = 35) - (-4x + 6z = 70)   37 37 Therefore X = -4 Next, we can find the other variables by subbing in ‘’x’’, for either equations 4 or 5. 11(-4) + 2z = -26  -44 + 44 + 2z = -26 +44  2z = 18  Z = 9 ……… . 2 2 So far, we have variables Z and X. Now we can go back to one of our original equations and sub in these values to find the final variable and solve the system. -9(-4) = 5y – 8(9) = -61  36 - 36 + 5y – 72 + 72 = -61 + 72 - 36  5y = -25  y = -5 Thus, we have solved the 3 variable system with a solution of (-4, -5, 9) Not hard when you think about it… just time consuming.
• 13. Linear Inequalities In mathematics , an inequality is a statement about the relative size or order of two objects.
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• 19. Rational Inequalities We are given this ; X – 2 (x -4)(x+6) < 0 How to solve this you ask? Four Simple Steps. FIRST We factor the Numerator and Denominator SECOND , Determine The CRITICAL Numbers THIRD , Place Critical Numbers on a Number Line LASTLY, Test each Interval **Note ; CRITICAL means the restricted values & the values that will make the numerator zero . Critical Values in this inequality are 2, 4 and -6 With those done, we can test values. *Test any value from each section  Testing -11  we get values of -  - = - Area / Testing 0  -  - = + Area - - + - + - Testing 3  +  + = - Area / Testing 11  + = + Area - + - + + **Note that the < in the equation means we’re looking for a negative area & vice versa. If something like X > 1 appears, just rearrange it to X - 1 > 0 (X-4) (X+3) (X-4) (X+3) Follow through Normally  X(X+3) – (X-4)  X ² +3X – X – 4  X ² + 2X – 4 (X-4) (X+3) (X-4) (X+3) (X-4) (X+3) Then, Just plot them on a number line and Test the area’s. So our solution is at ( - ∞ , -3)(4, ∞ ) So our solution is at ( - ∞ , -6)(2, 4 )
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