Pre Calculus 11 Mind Map #3

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Pre Calculus 11 Mind Map #3 by Mind Map: Pre Calculus 11 Mind Map #3

1. Inequalities

1.1. Linear

1.1.1. Graph the line

1.1.1.1. Shade the region that consists of all points that make the inequality true

1.1.1.2. 4x-3y<-12 -3y<-4x-12 y<4/3x+4

1.1.1.2.1. Graph as you would a normal linear function replacing the less than sign with an equal sign. Proceed to graph a dotted line using +4 as a y intercept.

1.2. Quadratic

1.2.1. One Variable

1.2.1.1. 2x^2+5x≤3 2x^2+5x-3≤0 FIND ROOTS 2x^2-1x+6x-3 x(2x-1)+3(2x-1) (x+3)(2x-1) x=-3 and 1/2 a) x≤-3 b) -3<x<1/2 c) x≥1/2 Test b) 0+0≤3 TRUE -3≤x≤1/2

1.2.2. 2 Variables

1.2.2.1. y≤-x^2+6x+9 (-6±√72)/(-2) ROOTS-.1243 and 7.243 Vertex=(-b)/2a p=3 q=18 Vertex=(3,18)

1.2.2.1.1. Graph the parabola given the roots and vertex. In this case the parabola would have a direction of opening downwards (sad). The inequality is less than or equal to so you would shade below the parabola which in this case would be inside the parabola.

1.2.2.2. y≥x^2+7x+6 (x+6)(x+1) Roots x=-6 and-1 Vertex (–b)/2a p= -7/2 p= -3.5 (-3.5)^2+7(-3.5)+6=q q=-6.25 Vertex=(-3.5,-6.25)

1.2.2.2.1. Graph the parabola given the roots and vertex. In this case the parabola would have a direction of opening upwards (happy). The inequality is greater than or equal to so you would shade above the parabola which in this case would be inside the parabola.

1.3. ≤=solid line < =dotted line

2. Solving Systems

2.1. Graphing

2.1.1. Plug the functions into a graphing calculator to find the points of intersection.

2.1.1.1. If the quadratic-quadratic or quadratic-linear have one intersecting point there is only one solution

2.1.1.2. If the quadratic-quadratic or quadratic-linear have two intersecting points there is only two solutions

2.1.1.3. If the quadratic-quadratic or quadratic-linear have no intersecting points there is no solutions

2.2. FINDING THE POINT(S) OF INTERSECTION

2.3. Algebra

2.3.1. Elimination

2.3.1.1. Eliminating one system from the other and solving

2.3.1.1.1. 3x^2-24x-y=-52 -2x^2-y+16x=28 3x^2-24x+52=y -2x^2+16x-28=y 5x^2-40x+80=0 5(x^2-8x+16)=0 (x-4)(x-4) 3(4)^2-24(4)+52=y y=4 (4,4)

2.3.2. Substitution

2.3.2.1. Substituting one system into the other and solving

2.3.2.1.1. x^2-x-6=y 2x-2=y x^2-x-6=2x-2 x^2-3x-4=0 (x-4)(x+1)=0 x=4 and 1 2(4)-2=y 6=y 2(-1)-2=y -4=y (4,6)(-1,-4)