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# Math 109 1st Midterm Exam Notes MA 109

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This 98 page Study Guide was uploaded by Javone Byrd on Tuesday August 30, 2016. The Study Guide belongs to MA 109 at University of Kentucky taught by Nicholas Nguyen in Fall 2016. Since its upload, it has received 20 views. For similar materials see College Algebra in Mathematics Education at University of Kentucky.

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College Algebra Quadratic Equations Nicholas Nguyen nicholas.nguyen@uk.eydu Department of Mathematics UK September 07, 2016 Agenda I Solving quadratic equations I The quadratic formula I Factoring using the quadratic formula’s roots I The discriminant: number of solutions I Completing the square I REEF questions Quadratic Equations A quadratic equation in x is any equation of the form ax +bx +c = 0; a = 6 0; or an equivalent equation. F OIL Reminder (a+b)▯(c +d) is equal to (expands to) ac + ad + bc + bd |{z} |{z} |{z} |{z} First Outer Inner Last Example: (x ▯1)(x +2) is equal to x ▯x + x ▯2 +(▯1)x +(▯1)(2) |{z} |{z} | {z } | {z } First Outer Inner Last 2 = x +2x ▯1x ▯2 = x +x ▯2: F actoring a quadratic is reversing this process. F OIL Reminder (a+b)▯(c +d) is equal to (expands to) ac + ad + bc + bd |{z} |{z} |{z} |{z} First Outer Inner Last Example: (x ▯1)(x +2) is equal to x ▯x + x ▯2 +(▯1)x +(▯1)(2) |{z} |{z} | {z } | {z } First Outer Inner Last 2 = x +2x ▯1x ▯2 = x +x ▯2: F actoring a quadratic is reversing this process. Zero Product Property Let A and B be quantities. If A▯B = 0, then either A = 0; or B = 0: Solving Using Factors Let A and B be quantities. If A▯B = 0, then either A = 0; or B = 0: Example: consider the equation (x ▯1)(x +2) = 0: Then either (x ▯1) = 0; or (x +2) = 0; so x ▯1 +1 = 0 +1; or x +2 ▯2 = 0 ▯2; x = 1; or x = ▯2 : What if the quadratic equation is not factored? Let’s nd solutions Solving Using Factors Let A and B be quantities. If A▯B = 0, then either A = 0; or B = 0: Example: consider the equation (x ▯1)(x +2) = 0: Then either (x ▯1) = 0; or (x +2) = 0; and we can solve these two equations: x ▯1 +1 = 0 +1; or x +2 ▯2 = 0 ▯2; x = 1; or x = ▯2 : What if the quadratic equation is not factored? Let’s nd solutions using a formula. Quadratic Formula The solutions of ax +bx +c = 0; where a =6 0, are ▯b▯ p b ▯4ac x = 2a where a is the coecient of x , b is the coecient of x, and c is the constant term. I oYu can only have x , x, and a constant term. T I All terms must be on the left. S I Put the terms in order: x rst, x second, constant third. The solutions of 2 ax +bx +c = 0; where a = 6 0, are p 2 ▯b▯ b ▯4ac x = 2a where a is the coecient of x , b is the coecient of x, and c is the constant term. I oYu can only have x , x, and a constant term. T I All terms must be on the left. A I Put the terms in order: x rst, x second, constant third. Quadratic Formula Example Solve for x: Move everything left. Add 2x 2 on both sides. 2 Get terms in order: x rst, x second, constant third. ▯x ▯3 = ▯2x 2 2 2 +2x +2x ▯x ▯3 +2x 2 = 0 2 2x ▯x ▯3 = 0 2x 2 ▯1x ▯3 = 0 a = 2; b = ▯1; c = ▯3 Quadratic Formula Example Solve for x: Move everything left. Add 2x 2 on both sides. 2 Get terms in order: x rst, x second, constant third. ▯x ▯3 = ▯2x 2 2 2 +2x +2x ▯x ▯3 +2x 2 = 0 2 2x ▯x ▯3 = 0 2x 2 ▯1x ▯3 = 0 a = 2; b = ▯1; c = ▯3 Quadratic Formula Example Solve for x: Move everything left. Add 2x 2 on both sides. 2 Get terms in order: x rst, x second, constant third. ▯x ▯3 = ▯2x 2 2 2 +2x +2x ▯x ▯3 +2x 2 = 0 2 2x ▯x ▯3 = 0 2x 2 ▯1x ▯3 = 0 a = 2; b = ▯1; c = ▯3 Quadratic Formula Example Solve for x: Move everything left. Add 2x 2 on both sides. 2 Get terms in order: x rst, x second, constant third. ▯x ▯3 = ▯2x 2 2 2 +2x +2x ▯x ▯3 +2x 2 = 0 2 2x ▯x ▯3 = 0 2x 2 ▯1x ▯3 = 0 a = 2; b = ▯1; c = ▯3 Quadratic Formula Example 2 Move everything left. Add 2x on both sides. Get terms in order: x rst, x second, constant third. 2 ▯x ▯3 = ▯2x +2x 2 +2x 2 ▯x ▯3 +2x 2 = 0 2 2x ▯x ▯3 = 0 2x 2 ▯1x ▯3 = 0 a = 2; b = ▯1; c = ▯3 p ▯b▯ b ▯4ac x = 2a p ▯(▯1) ▯ (▯1) 2 ▯4(2)(▯ 3) = 2(2) Quadratic Formula Example 2 Get terms in order: x rst, x second, constant third. 2 ▯x ▯3 +2x = 0 2x 2 ▯x ▯3 = 0 2 2x ▯1x ▯3 = 0 a = 2; b = ▯1; c = ▯3 p 2 x = ▯b▯ b ▯4ac 2a p 2 = ▯(▯1) ▯ (▯1) ▯4(2)(▯ 3) 2(2) p = 1 ▯ 1 +24 4 2 2x ▯x ▯3 = 0 2x2▯1x ▯3 = 0 a = 2; b = ▯1; c = ▯3 p ▯b▯ b ▯4ac x = 2ap ▯(▯1) ▯ (▯1) 2▯4(2)(▯ 3) = p 2(2) 1 ▯ 1 +24 = p4 1 ▯ 25 = 4 = 1 ▯5 4 a = 2; b = ▯1; c = ▯3 p 2 x = ▯b▯ b ▯4ac 2a p 2 = ▯(▯1) ▯ (▯1) ▯4(2)(▯ 3) 2(2) p = 1 ▯ 1 +24 4 p = 1 ▯ 25 4 1 ▯5 = 4 so 3 x = 1 +5 = 6 = ; o r x =1 ▯5 = ▯4 = ▯1 : 4 4 2 4 4 Quadratic Formula Check 2 ▯x ▯3 = ▯2x 3 x = ; o r x = ▯1 2 Check. Plug in x = 3=2: Plug in x = ▯1: 3 ▯ ▯3 2 ▯ ▯3 = ▯2 ▯(▯1) ▯3 = ▯2 (▯1) 2 2 2 ? 3 6 ▯ ▯9 1 ▯3 = ▯2 (1) ▯ ▯ = ▯2 2 2 4 ▯2 = ▯2X ▯9 ? ▯18 = 2 4 ▯9 ▯9 2 = 2 X Quadratic Formula Helps Us Factor Suppose we solve 2 ax +bx +c = 0 with the quadratic formula, giving us real solutions x = r ; x = r :1 2 Then we can factor 2 ax +bx +c = a▯(x ▯r )(x ▯r ): 1 2 Quadratic Formula Helps Us Factor We solved 2x 2 ▯x ▯3 = 0 and got 3 x = ; x = ▯1 : 2 Then we can factor ▯ ▯ 2x 2 ▯x ▯3 = 2 ▯ x ▯ 3 (x ▯(▯1)) 2 ▯ ▯ = 2 ▯ x ▯ 3 (x +1) 2 Quadratic Formula Helps Us Factor We solved 2x 2 ▯x ▯3 = 0 and got 3 x = ; x = ▯1 : 2 Then we can factor ▯ ▯ 2x 2 ▯x ▯3 = 2 ▯ x ▯ 3 (x ▯(▯1)) 2 ▯ ▯ = 2 ▯ x ▯ 3 (x +1) 2 F actoring Check Use FOIL: ▯ ▯ ▯ ▯ 3 3 3 2 ▯ x ▯ 2 (x +1) = 2 ▯ x ▯x +x ▯1 ▯ x ▯ 2 2 ▯1 ▯ ▯ 2 3 3 = 2 ▯ x +1x ▯ x ▯ 2 2 ▯ ▯ 2 2 3 3 = 2 ▯ x + 2 x ▯ x ▯2 2 ▯ ▯ 2 1 3 = 2 ▯ x ▯ x ▯ 2 2 2 1 3 = 2x ▯2 ▯ x 22 ▯ 2 2 = 2x ▯x ▯3X Numb er of Solutions In the quadratic formula p ▯b▯ b ▯4ac x = ; 2a 2 the expression b ▯4ac , called the discriminant , lets us predict how many real solutions ax +bx +c = 0 has: 2 l b ▯4ac Numb er of real solutions > 0 2 (distinct) = 0 1 (repeated) < 0 0 (can’t be factored) Using the Discriminant Find how many real solutions this equation has. Move everything left. Subtract 3 x from both sides. Get terms in order: x rst, x second, constant third. 3 +x +2x = 3x ▯3x ▯3x 3 +x ▯x 2 = 0 x ▯x +3 = 0 2 1x ▯1x +3 = 0 a = 1; b = ▯1; c = 3 b ▯4ac = (▯1) 2 ▯4(1)(3) = 1 ▯12 = ▯11 Using the Discriminant Find how many real solutions this equation has. Move everything left. Subtract 3 x from both sides. Get terms in order: x rst, x second, constant third. 3 +x +2x = 3x ▯3x ▯3x 3 +x ▯x 2 = 0 x ▯x +3 = 0 2 1x ▯1x +3 = 0 a = 1; b = ▯1; c = 3 b ▯4ac = (▯1) 2 ▯4(1)(3) = 1 ▯12 = ▯11 Using the Discriminant Find how many real solutions this equation has. Move everything left. Subtract 3 x from both sides. Get terms in order: x rst, x second, constant third. 3 +x +2x = 3x ▯3x ▯3x 3 +x ▯x 2 = 0 x ▯x +3 = 0 2 1x ▯1x +3 = 0 a = 1; b = ▯1; c = 3 b ▯4ac = (▯1) 2 ▯4(1)(3) = 1 ▯12 = ▯11 Using the Discriminant Find how many real solutions this equation has. Move everything left. Subtract 3 x from both sides. Get terms in order: x rst, x second, constant third. 3 +x +2x = 3x ▯3x ▯3x 3 +x ▯x 2 = 0 x ▯x +3 = 0 2 1x ▯1x +3 = 0 a = 1; b = ▯1; c = 3 b ▯4ac = (▯1) 2 ▯4(1)(3) = 1 ▯12 = ▯11 Using the Discriminant Find how many real solutions this equation has. Move everything left. Subtract 3 x from both sides. Get terms in order: x rst, x second, constant third. 3 +x +2x = 3x ▯3x ▯3x 3 +x ▯x 2 = 0 x ▯x +3 = 0 2 1x ▯1x +3 = 0 a = 1; b = ▯1; c = 3 b ▯4ac = (▯1) 2 ▯4(1)(3) = 1 ▯12 = ▯11 Using the Discriminant Find how many real solutions this equation has. Move everything left. Subtract 3 x from both sides. Get terms in order: x rst, x second, constant third. 3 +x +2x = 3x ▯3x ▯3x 3 +x ▯x 2 = 0 x ▯x +3 = 0 2 1x ▯1x +3 = 0 a = 1; b = ▯1; c = 3 b ▯4ac = (▯1) 2 ▯4(1)(3) = 1 ▯12 = ▯11 Find how many real solutions this equation has. 3 +x +2x = 3x ▯3x ▯3x 2 3 +x ▯x = 0 x ▯x +3 = 0 2 1x ▯1x +3 = 0 a = 1; b = ▯1; c = 3 2 2 b ▯4ac = (▯1) ▯4(1)(3) = 1 ▯12 = ▯11 < 0: The discriminant is negative. This quadratic equation has no real solutions and cannot be factored! P erfect Squares Expressions in the form ▯ ▯ 2 x +bx + b 2 can be factored as (are equal to) ▯ ▯ 2 x + b : 2 2 Identify b, then see if the constant term is (b=2) . These expressions are called perfect squares . Let us try factoring x +6x +9: b = 6, and the constant term is indeed ▯ ▯ 6 2 = (3) 2 = 9; 2 P erfect Squares Expressions in the form ▯ ▯ 2 x +bx + b 2 can be factored as (are equal to) ▯ ▯ 2 x + b : 2 2 Identify b, then see if the constant term is (b=2) . These expressions are called perfect squares . Let us try factoring x +6x +9: b = 6, and the constant term is indeed ▯ ▯ 6 2 = (3) 2 = 9; 2 P erfect Squares Expressions in the form ▯ ▯ 2 x +bx + b 2 can be factored as (are equal to) ▯ ▯ 2 x + b : 2 2 Identify b, then see if the constant term is (b=2) . These expressions are called perfect squares . Let us try factoring x +6x +9: b = 6, and the constant term is indeed ▯ ▯ 6 2 = (3) 2 = 9; 2 ▯ ▯ b 2 x 2 +b x + 2 can be factored as (are equal to) ▯ ▯ b 2 x + : 2 2 Identify b, then see if the constant term is (b=2) . These expressions are called perfect squares . Let us try factoring x +6x +9: b = 6, and the constant term is indeed ▯ ▯ 2 6 = (3) 2 = 9; 2 so ▯ ▯ 2 x +6x +9 = x + 6 = (x +3) : 2 2 Identify b, then see if the constant term is (b=2) 2 . These expressions are called perfect squares . Let us try factoring 2 x +6x +9: b = 6, and the constant term is indeed ▯ ▯ 6 2 2 = (3) = 9; 2 so ▯ ▯ 6 2 x +6x +9 = x + = (x +3) : 2 2 Check: (x +3) 2 = (x +3)(x +3) = x ▯x +x ▯3 +3 ▯x +3 ▯3 2 2 = x +3x +3x +3 = x +6x +9X P erfect Squares Example 2 2 x ▯8x +16 b = ▯8, and the constant term is indeed ▯ ▯ 2 ▯8 2 = (▯4) = 16; 2 so ▯ ▯8 ▯ 2 x ▯8x +16 = x + = (x ▯4) : 2 2 Check: 2 (x ▯4) = (x ▯4)(x ▯4) = x ▯x +x ▯(▯4) ▯4 ▯x +(▯4) ▯(▯4) = x ▯4x ▯4x +(▯4) 2 2 = x ▯8x +16X P erfect Squares Example 3 2 81 x +9x + 4 b = 9, and the constant term is indeed ▯ ▯ 9 2 81 = 2 4 so ▯ ▯ 2 2 81 9 x +9x + 4 = x + 2 ▯ ▯ 81 9 2 x +9x + = x + 4 2 Check: ▯ ▯ 2 ▯ ▯ ▯ ▯ 9 9 9 x + 2 = x + 2 x + 2 9 9 9 9 = x ▯x +x ▯ 2 + 2 ▯x + 2 ▯ 2 ▯ ▯ 2 2 9 9 9 = x + 2 x + x 2 2 = x + 2 18 x + 81 2 4 81 = x +9x + X 4 Completing the Square If you have x +bx +::: = 0 in your quadratic equation, you can get a power equation by completing the square: Move constants to right, x s to left 2 Divide every term by the coecient of x Add (b=2) 2 to both sides Factor the perfect square (on left side) Let us try this on 2 2x +20x = ▯6: Completing the Square If you have x +bx +::: = 0 in your quadratic equation, you can get a power equation by completing the square: Move constants to right, x s to left 2 Divide every term by the coecient of x Add (b=2) 2 to both sides Factor the perfect square (on left side) Let us try this on 2 2x +20x = ▯6: If you have 2 x +bx +::: = 0 in your quadratic equation, you can get a power equation by completing the square: Move constants to right, x s to left Divide every term by the coecient of x 2 2 Add (b=2) to both sides Factor the perfect square (on left side) Let us try this on 2 2x +20x = ▯6: Can’t complete the square yet. Divide by 2: 2x 2 20x ▯6 + = 2 2 2 2 x +10x = ▯3 On left, b = 10, 2 Divide every term 2y the coecient of x Add (b=2) to both sides Factor the perfect square (on left side) Let us try this on 2x 2 +20x = ▯6: Can’t complete the square yet. Divide by 2: 2 2x + 20x = ▯6 2 2 2 2 x +10x = ▯3 On left, b = 10, so add ▯ ▯ 10 2 = (5) 2 2 to both sides: x +10x +(5) 2 = ▯3 +(5) 2 Let us try this on 2 2x +20x = ▯6: Can’t complete the square yet. Divide by 2: 2x 2 20x ▯6 + = 2 2 2 x +10x = ▯3 On left, b = 10, so add ▯ ▯ 2 10 2 2 = (5) to both sides: x +10x +(5) 2 = ▯3 +(5) 2 Factor left side: (x +5) 2 = 22 Can’t complete the square yet. Divide by 2: 2x 2 20x ▯6 + = 2 2 2 x +10x = ▯3 On left, b = 10, so add ▯ ▯ 2 10 2 2 = (5) to both sides: 2 2 2 x +10x +(5) = ▯3 +(5) Factor left side: (x +5) 2 = 22 Power equation: Even power, non-negative right side, so 2 roots: p p x +5 = 22; x +5 = ▯ 22; On left, b = 10, so add ▯ ▯ 10 2 2 = (5) 2 to both sides: x +10x +(5) 2 = ▯3 +(5) 2 Factor left side: (x +5) 2 = 22 Power equation: Even power, non-negative right side, so 2 roots: p p x +5 = 22; x +5 = ▯ 22; so p p x = ▯5 + 22; x = ▯5 ▯ 22 : C1: What is a? We want to solve the quadratic equation 5x 2 +8x +12 = ▯29x 2 ; using the quadratic equation with b = 8; c = 12: What number should a equal for the quadratic equation? yTpe in a whole number. C1: What is a? We want to solve the quadratic equation 2 2 5x +8x +12 = ▯29x ; using the quadratic equation with b = 8; c = 12: What number should a equal for the quadratic equation? Move everything left: 5x 2 +8x +12 = ▯29x 2 2 2 +29x +8x +12: +29x 34x 2 +8x +12 = 0 The number a is the coecient of x : 34 . C2: How Many Solutions? Use the discriminant to predict ho w many solutions 2x 2 ▯8x +8 = 0 has. T ype in the number of solutions this equation has. C2: How Many Solutions? Use the discriminant to predict ho w many solutions 2x 2 ▯8x +8 = 0 has. a = 2; b = ▯8; c = 8 The discriminant is 2 2 b ▯4ac = (▯8) ▯4(2)(8) = 64 ▯64 = 0: Since the discriminant is zero , the equation has one (repeated) solution . C3: Completing the Square In the equation, 2 x +14x = 6; we want to complete the square. What number should we add on both sides? Hint: the constant is already on the right, and the x’s are on the left. Which coecient should you look at? C3: Completing the Square In the equation, x +14x = 6; we want to complete the square. What number should we add on both sides? On the left, b, the coecient of x, is b = 14; so we need to add ▯ ▯ 2 ▯ ▯ 2 b = 14 = 7 2 = 49 2 2 to both sides. C3: Completing the Square Finding the solutions to the equation so we need to add ▯ ▯ 2 ▯ ▯ 2 b = 14 = 7 2 = 49 2 2 to both sides. x +14x = 6 +7 2 +49 2 2 x +14x +7 = 55 (x +7) 2 = 55 Even power, non-negative right side, so two roots: p p x +7 = 55; x +7 = ▯ 55 so the solutions are p p x = ▯7 + 55; x = ▯7 ▯ 55: Next Time I Please read Section 1.2 of your textbook. I We will study equations of quadratic type, solving by factoring, and equations with square roots. College Algebra Equations of Lines Nicholas Nguyen nicholas.nguyen@uk.eydu Department of Mathematics UK September 16, 2016 Agenda I P oint-slope form I Comparison of forms I Parallel and perpendicular lines I REEF questions Slop e I The slope of the line segment between (x ;1 ) an1 (x ;y ), where x 6= x , is 2 2 1 2 y 2y 1 rise ▯y = = x 2x 1 run ▯x If you move right x 2x units 1n the line, you go up y 2y units.1 I If x =1x , the 2ine segment is vertical, and its slope is undened (cannot go left or right on a vertical line). Example: The slope of the line segment from (4; 7) to (8; 9) is 9 ▯7 2 1 = = : 8 ▯4 4 2 Slop e over a Run of 4 Example: The slope of the line segment from (4; 7) to (8; 9) is 9 ▯7 = 2 = 1 : 8 ▯4 4 2 If you go 4 units from x = 4 to x = 8 on the line, then your y-coordinate rises by 2 units (half of 4) from 7 to 9. Slop e Over a Run of 1 Example: The slope of the line segment from (4; 7) to (8; 9) is 9 ▯7 = 2 = 1 : 8 ▯4 4 2 If you go 1 unit from x = 4 to x = 5 on the line, then your y-coordinate rises by 1 =2 from 7 to 7.5. P oint-Slope Form of a Line An equation in point-slope form for a line with slope m that goes through the point (a;b) is y ▯b = m(x ▯a) Why? Let (x;y) be a point on the line. Then the slope between (a;b) and (x;y) is y ▯b ; x ▯a but since the line has slope m, this fraction must equal m: y ▯b = m: x ▯a Ho wever, we don’t want x in a denominator, so multiply both sides by x ▯a: (y ▯b) (x ▯a) (x ▯a) = m(x ▯a) An equation in point-slope form for a line with slope m that goes through the point (a;b) is y ▯b = m(x ▯a) Why? Let (x;y) be a point on the line. Then the slope between (a;b) and (x;y) is y ▯b ; x ▯a but since the line has slope m, this fraction must equal m: y ▯b x ▯a = m: Ho wever, we don’t want x in a denominator, so multiply both sides by x ▯a: (x ▯a) (y ▯b) = m(x ▯a) (x ▯a) y ▯b = m(x ▯a) Why? Let (x;y) be a point on the line. Then the slope between (a;b) and (x;y) is y ▯b ; x ▯a but since the line has slope m, this fraction must equal m: y ▯b = m: x ▯a Ho wever, we don’t want x in a denominator, so multiply both sides by x ▯a: (y ▯b) (x ▯a) (x ▯a) = m(x ▯a) y ▯b = m(x ▯a) I An equation of a line with slope ▯3 through (▯4; 5) is y ▯5 = (▯3)(x ▯(▯4)) y ▯5 = ▯3(x +4): but since the line has slope m, this fraction must equal m: y ▯b x ▯a = m: Ho wever, we don’t want x in a denominator, so multiply both sides by x ▯a: (x ▯a) (y ▯b) = m(x ▯a) (x ▯a) y ▯b = m(x ▯a) I An equation of a line with slope ▯3 through (▯4; 5) is y ▯5 = (▯3)(x ▯(▯4)) y ▯5 = ▯3(x +4): I The line with equation y +1 = 2(x +3) Ho wever, we don’t want x in a denominator, so multiply both sides by x ▯a: (y ▯b) (x ▯a) = m(x ▯a) (x ▯a) y ▯b = m(x ▯a) I An equation of a line with slope ▯3 through (▯4; 5) is y ▯5 = (▯3)(x ▯(▯4)) y ▯5 = ▯3(x +4): I The line with equation y +1 = 2(x +3) y ▯(▯1) = 2(x ▯(▯3)) is a line with slope 2 through (▯3; ▯1). y ▯b = m(x ▯a) I An equation of a line with slope ▯3 through (▯4; 5) is y ▯5 = (▯3)(x ▯(▯4)) y ▯5 = ▯3(x +4): I The line with equation y +1 = 2(x +3) y ▯(▯1) = 2(x ▯(▯3)) is a line with slope 2 through (▯3; ▯1). I An equation of a line with slope 0 through (6; 7) is y ▯7 = 0(x ▯6) I An equation of a line with slope ▯3 through (▯4; 5) is y ▯5 = (▯3)(x ▯(▯4)) y ▯5 = ▯3(x +4): I The line with equation y +1 = 2(x +3) y ▯(▯1) = 2(x ▯(▯3)) is a line with slope 2 through (▯3; ▯1). I An equation of a line with slope 0 through (6; 7) is y ▯7 = 0(x ▯6) y ▯7 = 0 y = 7 This is a horizontal line whose points all have y-coordinate 7. Pros/Cons: Point-slope Form Line has slope m and goes through (a;b): y ▯b = m(x ▯a) Pro: I Often, you are given the slope and a point, so you can write this form down. This is the best form to start with. Cons: I Not unique, but you can convert to slope-intercept form: solve for y, then distribute on right. I Cannot handle vertical lines (slope is undened) Pros/Cons: Slope-intercept Form Line has slope m and goes through (0; b), so y-intercept is b: y = mx +b Pro: I Unique - good as a nal answer after converting from another form. Cons: I Often, you are not given the y-intercept. Use point-slope form rst. I Cannot handle vertical lines Pros/Cons: General Linear Form A, B, and C are coecients or constants: Ax +By +C = 0 Pro: I Can handle any line: horizontal, vertical, and anything else. Con: I Must convert to slope-intercept form to nd slope, unless B = 0, in which case solve for x to get a vertical line. Ho rizontal/Vertical Lines The graph of y = b is a horizontal line that passes through (a;b). I The slope of a horizontal line is 0. The graph of x = a is a vertical line that passes through (a;b). I The slope of a vertical line is undened. Example: Line Through Two Points Find an equation for a line through (▯1; ▯2) and (3; 5) in slope-intercept form. I Find point-slope form rst: Need point (two choices) and slope (use points) I Convert to slope-intercept form: Solve for y and distribute Line Through Two Points: Point-slope Form Find an equation for a line through (▯1; ▯2) and (3; 5) in slope-intercept form. I Find slope between points: 5 ▯(▯2) = 5 +2 = : 7 3 ▯(▯1) 3 +1 4 I W rite point-slope form using (3; 5): 7 y ▯5 = (x ▯3) 4 I Convert to slope-intercept form Solve for y: 7 y = (x ▯4) +5 Distribute: y = 7 x ▯3 ▯ 7 +5 Line Through Two Points: Point-slope Form Find an equation for a line through (▯1; ▯2) and (3; 5) in slope-intercept form. I Find slope between points: 5 ▯(▯2) = 5 +2 = : 7 3 ▯(▯1) 3 +1 4 I W rite point-slope form using (3; 5): 7 y ▯5 = (x ▯3) 4 I Convert to slope-intercept form Solve for y: 7 y = (x ▯3) +5 4 Distribute: 7 7 y = 4 x ▯3 ▯ 4 +5 Line Through Two Points: Point-slope Form Find an equation for a line through (▯1; ▯2) and (3; 5) in slope-intercept form. I Find slope between points: 5 ▯(▯2) = 5 +2 = : 7 3 ▯(▯1) 3 +1 4 I W rite point-slope form using (3; 5): 7 y ▯5 = (x ▯3) 4 I Convert to slope-intercept form Solve for y: 7 y = (x ▯3) +5 4 Distribute: 7 7 y = 4 x ▯3 ▯ 4 +5 I W rite point-slope form using (3; 5): 7 y ▯5 = (x ▯3) 4 I Convert to slope-intercept form Solve for y: 7 y = (x ▯3) +5 4 Distribute: 7 7 y = x ▯3 ▯ +5 4 4 7 21 5 y = 4 x ▯ 4 + 1 7 21 20 y = x ▯ + 4 4 4 7 1 y = 4 x ▯ : 4 The slope is 7=4 and the y-intercept is ▯1=4. Line Through Two Points: Point-slope Form (alt) Find an equation for a line through (▯1; ▯2) and (3; 5) in slope-intercept form. I Find slope between points: 5 ▯(▯2) 5 +2 7 = = : 3 ▯(▯1) 3 +1 4 I W rite point-slope form using (▯1; ▯2): 7 y ▯(▯2) = 4 (x ▯(▯1)) y +2 = 7 (x +1) 4 7 y +2 = (x +1) 4 I Convert to slope-intercept form Solve for y: 7 y = 4 (x +1) ▯2 Distribute: 7 7 y = 4 x +1 ▯ 4 ▯2 7 7 2 y = x + ▯ 4 4 1 7 7 8 y = 4 x + 4 ▯ 4 y = 7 x ▯ : 1 4 4 The slope is 7 =4 and the y-intercept is ▯1=4. This is the same equation we got earlier when we used the point (3; 5). Uniqueness of Slope-Intercept Form A line can be described by many dierent equations in point-slope form by choosing dierent points, but a line only has one equation in slope-intercept form because a line (that is not vertical) can only have one y-intercept. P arallel and Perpendicular Lines Suppose we have a given line whose slope is m. I If a second line is parallel to this given line, then the slope of the second line is m = slope of given line: I If a second line is perpendicular to this given line, then the slope of the second line is 1 ▯ m = negative reciprocal of slope of given line. P arallel and Perpendicular Lines Suppose we have a given line whose slope is m. I If a second line is parallel to this given line, then the slope of the second line is m = slope of given line: I If a second line is perpendicular to this given line, then the slope of the second line is 1 ▯ m = negative reciprocal of slope of given line. P erpendicular Line Example Find an equation for the line perpendicular to 4 x ▯3y ▯6 = 0 and goes through (5; 7). I Need slope - nd rst line’s slope by solving for y: ▯3y = ▯4x +6 ▯3y ▯4x 6 = + ▯3 ▯3 ▯3 4 y = 3 x ▯2 The rst line has slope 4 =3, so the slope of the line we want is the negative reciprocal: ▯ 3 4 P erpendicular Line Example Find an equation for the line perpendicular to 4 x ▯3y ▯6 = 0 and goes through (5; 7). I Need slope - nd rst line’s slope by solving for y: ▯3y = ▯4x +6 ▯3y ▯4x 6 = + ▯3 ▯3 ▯3 4 y = 3 x ▯2 The rst line has slope 4 =3, so the slope of the line we want is the negative reciprocal: ▯ 3 4 P erpendicular Line Example Find an equation for the line perpendicular to 4 x ▯3y ▯6 = 0 and goes through (5; 7). I Need slope - nd rst line’s slope by solving for

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