# Week 15 – Precalc 11

This week in Precalculus 11, we learned about solving rational equations.

Rational equation: an equation containing rational expressions.

To solve a rational equation, we first identify non-permissible values. One way to solve a rational equation is to multiply each term by the lowest common denominator, then solve normally. Another is if 2 fractions are equal to eachother and have either the same numerators or same denominators, their numerators/denominators must also match. We then check for extraneous roots, or non-permissible values.

Example:

# Week 14 – Precalc 11

This week in Precalculus 11, we started the Rational Expressions and Equations Unit. We first learned about equivalent rational expressions.

rational expression: a fraction with polynomials in the numerator & denominator

non-permissable value: variable values making the denominator 0

To determine the non-permissable values, we equate the denominator to 0 and solve the equation.

Example:

To simplify a rational expression, we first factor the numerator and denominator. We then identify the non-permissable values. Last, we cancel out identical pairs of numerators and denominators.

Example:

# Week 13 – Precalc 11

This week in Precalculus 11, we learned about graphing reciprocals of linear functions.

reciprocal: numerator and denominator are switched

linear function: an equation in 2 variables with degree 1

asymptote: barrier lines of a graph

To graph the reciprocal of a linear function, we first graph its parent function. We then find the points on the graph where y = 1 and y = -1, and the vertical and horizontal asymptotes. In most cases, the horizontal asymptote is y = 0. The vertical asymptote is the x-intercept. We then use these points to draw a 2-part graph known as a hyperbola. The asymptotes act as barriers.

Example:

# Week 12 – Precalc 11

This week in Precalculus 11, we started the Absolute Value and Reciprocal Functions unit. We first learned about absolute value functions.

Absolute value function: y = |f(x)|

Critical points: an absolute value function’s x-intercepts

An absolute value function’s graph changes direction at the critical points. To graph an absolute value function, we first graph its equivalent normal function [y = f(x)]. Any points below the x-axis are reflected along the x-axis, since absolute values are only positive.

Example:

An absolute value function can be written in piecewise notation. To do so, we write one function in which the absolute value expression is positive or 0, and one function in which the absolute value expression is negative. We then state when this occurs, using </>/≥/≤ and the critical point.

Example:

# Week 11 – Precalc 11

This week in Precalculus 11, we learned about graphing linear inequalities in 2 variables.

linear inequality: written in general form as ax + by + c </>/≤/≥ 0

To graph a linear inequality in 2 variables, we first graph its corresponding linear equation. To do this, we must rearrange the equation to standard form, where y = mx + b.

We then plot the y-intercept (b) and use the slope (m) to plot more points. We connect these points with a straight line. If the symbol is < or >, it means less than or greater than, and we use a broken line. If the symbol is ≤ or ≥, it means less than or equal to or greater than or equal to, and we use a solid line.

One side of the line has solutions, and the other does not. To determine which side does, we plug a test point (labelled T) into the original equation. (0, 0) is a good test point. If the statement is true, we shade in the side of the graph with said point. If not, we shade in the other side.

This was interesting, as I had learned how to graph linear equations, but not inequalities.

Example:

# Week 10 – Precalc 11

This week in Precalculus 11, we started the Graphing Inequalities & Systems of Equations unit. We first learned about solving quadratic inequalities in one variable.

quadratic inequality: an expression with degree 2 that is not equal to 0

It can be written in general form as a$x^2$ + bx + c >/</≥/≤ 0

To solve a quadratic inequality, we first rearrange the inequality to general form. We then write and solve its corresponding quadratic equation. We label a number line with the roots. We then test a number from each interval by substituting it into the original inequality.

Example:

This was new to me, as I had learned how to solve linear inequalities, but not quadratic inequalities.

# Week 9 – Precalc 11

This week in Precalculus 11, we learned about equivalent forms of a quadratic function.

x-intercept: point on a graph crossing the x-axis

root: x-value to an equation when solved

y-intercept: point on a graph crossing the y-axis

vertex: maximum (opening down) or minimum (opening up) point on a parabola

A quadratic function has 3 forms: factored, general, and standard.

factored form: y = a(x – $x_1$)(x – $x_2$)

• $x_1$ & $x_2$ are the graph’s x-intercepts, or the equation’s roots

general form: y = a$x^2$ + bx + c

• c is the graph’s y-intercept

standard form: y = a$(x - p)^2$ + q

• p is the vertex’s x-value
• q is the vertex’s y-value

For all 3 forms, a determines whether the parabola opens up (positive) or down (negative) and its stretch (a>1) or compression (a<1) value.

By converting from one form to another, we can determine all elements of the equation.

# Week 8 – Precalc 11

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$x^2$ + 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 $x^2$ 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$x^2$ + 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.

# Week 7 – Precalc 11

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

Real root: square root of a positive number

In the quadratic formula, the discriminant is $b^2$ – 4ac. By solving the discriminant, we can indicate how many real roots the equation has.

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

Example: 5$x^2$ – 9x + 4 = 0

$b^2$ – 4ac

= $(-9)^2$ – 4(5)(4)

= 81 – 20 · 4

= 81 – 80

= 1 -> 2 real roots

If $b^2$ – 4ac = 0, the equation has 1 real root.

Example: 2$x^2$ + 16x + 32 = 0

$b^2$ – 4ac

= $(16)^2$ – 4(2)(32)

= 256 – 8 · 32

= 256 – 256

= 0 -> 1 real root

If $b^2$ – 4ac < 0, the equation has 0 real roots.

Example: 6$x^2$ + 7 = 0

$b^2$ – 4ac

= $(0)^2$ – 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.

# Week 6 – Precalc 11

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:

$x^2$ – 6x = 27

$x^2$ – 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.

$x^2$ – 6x = 27

$(9)^2$ – 6(9) = 27

81 – 54 = 27

27 = 27

OR

$x^2$ – 6x = 27

$(-3)^2$ – 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.