This week in Precalculus 11, we started the Analyzing Quadratic Functions unit. We first learned about the properties of a quadratic function.

quadratic function: function that can be written in general form

general form: y = a + bx + c, where a ≠ 0

parabola: the curve of a quadratic function’s graph

vertex: a parabola’s highest (minimum) or lowest (maximum) point (if the coefficient of is positive, the vertex opens up. If it is negative, the vertex opens down)

axis of symmetry: intersects parabola at its vertex

domain: all possible x values

range: all possible y values

x-intercepts: where the parabola touches the x axis, the roots of the quadratic equation

y-intercepts: where the parabola touches the y axis.

 

Example: y = 2 + 8x + 6

x: -5, -4, -3, -2, -1, 0

y: 16, 6, 0, -2, 0, 6

vertex: (-2, -2) minimum/opens up

x-intercept: (-1, 0) & (-3, 0)

y-intercept: (0, 6)

axis of symmetry: x = -2

D: {x∈R}

R: {y ≥ -2}

 

I had not learned about many of these terms before and found it interesting that you could find so much information from the vertex.

This week in Precalculus 11, we learned about interpreting the discriminant.

Discriminant/Radicand: the number underneath a radical sign (√)

Real root: square root of a positive number

In the quadratic formula, the discriminant is – 4ac. By solving the discriminant, we can indicate how many real roots the equation has.

If – 4ac > 0, the equation has 2 real roots.

Example: 5 – 9x + 4 = 0

– 4ac

= – 4(5)(4)

= 81 – 20 · 4

= 81 – 80

= 1 -> 2 real roots

If  – 4ac = 0, the equation has 1 real root.

Example: 2 + 16x + 32 = 0

– 4ac

= – 4(2)(32)

= 256 – 8 · 32

= 256 – 256

= 0 -> 1 real root

If  – 4ac < 0, the equation has 0 real roots.

Example: 6 + 7 = 0

– 4ac

= – 4(6)(7)

= 0 – 24 · 7

= 0 – 168

= -168 -> 0 real roots

By interpreting the discriminant beforehand, we can determine whether or not to bother solving a quadratic equation.

 

 

This week in Precalculus 11, we learned about solving quadratic equations by factoring.

Solving: finding a variable’s value

Quadratic equation: an equation with a degree of 2

Degree: sum of the variables’ exponents in a term

Factoring: separating an expression into its components

Variable: an unknown number represented by a letter

The zero-product property is one strategy to solve quadratic equations. If the product of two numbers is 0, than one or both of the numbers have a value of 0. To solve a quadratic equation using this strategy, we first move all terms to one side of the equation, so the other has a value of 0. 

Example:

– 6x = 27

– 6x – 27 = 0

We then factor the expression.

(x – 9)(x + 3) = 0

Then we make one of the components have a value of zero and solve for the variable.

x – 9 = 0

x = 0 + 9

x = 9

OR

x + 3 = 0

x = 0 – 3

x = -3

We then verify the solutions by replacing x with the solutions in the original equation.

– 6x = 27

– 6(9) = 27

81 – 54 = 27

27 = 27

OR

– 6x = 27

– 6(-3) = 27

9 + 18 = 27

27 = 27

This is an interesting new way to solve equations, especially since I did not know a variable could have more than one solution. 

 

 

This week in Precalculus 11, we started the Solving Quadratic Equations Unit. We first reviewed factoring polynomial expressions.

Factoring: separating an expression into its components

Polynomial expression: an expression of numbers and variables being added, subtracted or multiplied (example: 2x)

Greatest common factor: the greatest number that can divide into all the terms in the expression (example: 2, 4, 6, 8, 10      GCF = 2)

Term: a number or a group of numbers being multiplied (example: 2)

Binomial: a polynomial expression with 2 terms (example: x + 1)

Difference of squares: a polynomial expression in which subtraction takes place between 2 perfect square terms (example: 4x – 1)

Trinomial: a polynomial expression with 3 terms (example: + x + 1)

Conjugates: 2 terms with opposite addition/subtraction signs (example: 1 + 1 & 1 – 1)

To factor a polynomial expression, we first look for the greatest common factor between the terms and divide each term by that number.

Example:

2x + 2

= 2(x + 1)

If the expression is a binomial, we check if it is a difference of squares. When factored, a difference of squares results in conjugates.

Example:

4x – 1

= (2x + 1)(2x – 1)

If the expression is a trinomial, we check if it is in the form a + bx + c. If a = 1, we separate bx into 2 terms that multiply to c and add to bx. 

Example:

+ 2x + 1

= (x + 1)(x + 1)

=

If a ≠ 1, we separate bx into 2 terms that multiply to ac and add to bx, then find the greatest common factor of each side, and divide each term by that number.

Example:

2 + 4x + 2

= 2 + 2x + 2x + 2

= 2x(x + 1) + 2(x + 1)

= (2x + 2)(x + 1)

If the expression is in a different form than a + bx + c, the expression can sometimes be changed to this form.

Example:

+ 2(x + 1) + 1

= 2a + 1

= (a + 1)(a + 1)

=

=

=