2. Square Root Function
• Basic Equation:
• y = √x
• Transformed
Equation: y = a√bx
• E.g. y = 2 √4x
• y = 2*2√x
• y = 4√x
x y = √x y = 2√4x
y = 2*2√x
0 0 4*0 = 0
1 1 4*1 = 4
4 2 4*2 = 8
9 3 4*3 = 12
3. Square Root Function
• Basic Equation:
• y = √x
• Transformed
Equation: y = a√bx
• E.g. y = 2 √-4x
• y = 2*2√-x
• y = 4√-x
x y = √x y = 2√-4x
y = 2*2√-x
0 0 4*0 = 0
1 1 4*1 = -4
4 2 4*2 = -8
9 3 4*3 = -12
4. Determining the Equation
• Basic Equation:
• y = √x
• Transformed
Equation: y = a√bx
• Substitute
• 10 = a√4
• 10 = a * 2
• 10/2 = 5 = a
• So y = 5√x
x y = √x y = a√bx?
0 0 0
1 1 5
4 2 10
9 3 15
5. Solving a Square Root Equation
• You may be asked to determine when the
equation from the last slide = 40
• y = 5√x
• 40 = 5√x
• Divide both sides by 5
• 8 = √x
• x = 2.8
• So when x = 2.8, y = 40
6. Inequalities and
Square Root Functions
• To solve inequalities, we treat them like
equations.
• When is 5√x < 50?
• Divide both sides by 5
• √x < 10
• Square both sides to get rid of square root
• x < 100
• The point x would NOT be part of the solution.
7. Quadratic or 2nd Degree Function
• Basic Equation:
• y = x2
• Transformed
Equation: y = ax2
• E.g. y = 2 (3x)2
• y = 2 * 9x2
• y = 18x2
x y = x2 y = 18x2
0 0 18*0 = 0
1 1 18*1 = 18
3 9 18*9 = 162
5 25 18*25 = 900
8. Determining the Quadratic Equation
• Basic Equation:
• y = x2
• Transformed
Equation: y = ax2
• Substitute
• 28 = a *22
• 28 = a * 4
• 28/4 = 7 = a
• So y = 7x2
x y = x2 y = ax2
0 0 0
1 1 7
2 4 28
3 9 63
9. Solving a Quadratic Equation
• You may be asked to determine when the
equation from the last slide = 400
• y = 18x2
• 400 = 18x2
• Divide both sides by 18
• 22.22 = x2
• x = ±4.71, positive and negative!
• So when x =± 4.71, y = 400
10. Inequalities and
Quadratic Functions
• To graph inequalities, we treat them like
equations.
• y ≥ 7 x2
• We would draw the graph y = 7 x2
• Because the equation is greater than AND
equal to, we shade above the line and
make the line SOLID.
12. Substituting
• y > x2
• Sub a point (0,1) in
and see if the
mathematical
equation is true or
not.
y > x2
1 (0)2
1 0
True, therefore
(0,1) is in the
region y>x2
14. Substituting
• y < x2
• Sub a point (2,0) in
and see if the
mathematical
equation is true or
not.
y < x2
0 (2)2
0 4
True, therefore
(2,0) is in the
region y<x2