Chapter 9 Quadratic Equations And Functions By Chris Posey and Chris Bell

9-1 Square Roots You already know how to find the square of a number i.e. 4²=16 To find the square root, you basically do the opposite of squaring a number. i.e. √(16)=4

Apply Skills Learned  Now that you know how find the square root, let’s try some problems for practice! Find the square roots of the following problems. 1. √(144)2. √(16)3. √(400)4 √(36) 4. √36=6 3. √400=20 2. √16=4 1. √122=12

9-2 Solving by Quadratic Equations by Finding Square Roots Example 1> x 2 =4 Write the original equation x= + √4 or - √4 Find square roots x= 2 or =4 and (-2 2 )=4 Example 2> 3x 2 -48=0 Write the original equation 3x 2 =48 Add 48 to each side x 2 =16 Divide each side by 3 x= √16 Find square roots x=4 or =16 and (-4) 2 =16

Apply Skills Learned Now that you know how to solve quadratic equations by finding square roots, find the answers to the following problems. 1.5t =02. x 2 =2253. x 2 -15=10 3. x=5 2. x=15 1. t=5

9.3 Simplifying Radicals Product Property of Radicals: √ab= √a ∙ √b Example> √50= √(25∙2)= 5 √2 Example> √48= √(4∙12)= 4 √3 Quotient Property of Radicals: √a/b= √a/√b Example> √32/50= √(2∙16)/√(2∙25) Factor using square roots =√(16/25) Divide common factors =√(16)/√(25) Use Quotient Property =4/5 Simplify

Apply Skills Learned Now that you know how to simplify radicals, try it out on your own! 1.√(9/49)  √(18) 3. √(196)  √2 3)14

9.4 Graphing Quadratic Functions A quadratic function is a function that can be written as a formula Y=ax 2 +bx+c, where a≠0 This will give a graph with a u shape called a parabola. If a is grater than 0, it opens up. If it is negative, then it opens down.

Graphing (9.4) Find the x coordinate of the vertex, which is x=-b/2a Make a x,y table and use the x values Plot the points and connect them with a smooth curve to form the parabola

Example Sketch graph of y=x 2 -2x-3 Find x coordinate of vertex. -b/2a=-2/2(1)=1 Make a table x| y| (1,-4) is the vertex, plot the rest of the points and draw a curve connecting them.

Practice Find vertex coordinates, and make a x,y value table using x values to the right and left of the vertex. y=-4x 2 -4x+8  x| y|

9.5 Solving Quadratic Equations by Graphing Solutions for the quadratic graphs are the x-axis intercepts, where y=0. This number can be checked in the original equation, by setting it equal to 0.

Example y=x 2 -2x-3 is shown here. Note that the x intercepts are located at -1 and 3. If substituted for x, the equation would result in zero, the solutions for the equation.

Practice Solve the equation algebraically, check your answers by graphing. 2x 2 +8=16 Solutions are ±2

9.6 Solving Quadratic Equations by the Quadratic Formula The solutions of the quadratic equation, ax 2 +by+c=0, are (-b+/-√(b 2 -4ac) x= (2a) when a≠0 and b 2 -4ac≥0

Example Solve x 2 +9x=14=0 Solution 1x+9x+14=0 (-b+/-√(b 2 -4ac) x= (2a) (-(9)+/-√((9) 2 -4(1)(14)) x= (2(1)) -9+/-√(25) x=

-9+/-5 X= There are 2 solutions x=-2, and x=-7

Practice Solve the quadratic equation. 2x 2 -3x=8 X=2.89, and x=-1.39

9.7 Using the Discriminant The discriminant is the radical expression in the quadratic formula ie. (-b+/-√(b 2 -4ac)) x= (2a) If the discriminant is positive, then the solution has 2 solutions. If it is zero, it has one solution. If it is negative, there are no real solutions.

Example Find value of the discriminat and determine if it has two solutions, one solution, or no solutions. x 2 -3x-4=0 Use the equation, ax 2 +bx+c=0 to identify values, ie. a=1, b=-3, c=-4 Substitute into the discriminant b 2 -4ac=(-3) 2 -4(1)(-4) =9+16 =25 Discriminant is positive, therefore two solutions.

Practice Determine whether the graph will intersect the x-axis at one, two, or zero points. y=x 2 -2x+4 It does not intersect the axis

9.8 Graphing Quadratic Inequalities The graph of a quadratic inequality consists of all the points (x,y) that are part of the inequality. Quadratic inequalities can be represented by; y> or < or  or  ax 2 +bx+c

When graphing, use a dashed line for the parabola when the equality is > or< Use a solid line when it is  or  The parabola separates the graph into two sections. A test point is a point that is not on the graph. If the test point is a solution, then shade the region, if not shade the other region.

Example Solve –x < 0. Find the x-axis intercepts –x = 0 x2 – 4 = 0 (x + 2)(x – 2) = 0 x = –2 or x = 2 Use the origional inequality to find the area to shade y<0, therefore shade everything outside the parabola, below the x-axis.

Practice Determine whether the orderd pair is a solution of the inequality. y≤x 2 +7, (4,31) Point is a solution outside of the parabola

Now remember to study hard, because we’ll be watching you….