Week of March 12, 2018

Algebra 1

Often, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring. While factoring may not always be successful, the Quadratic Formula can always find the solution.

The Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve

The Quadratic Formula is derived from the process of completing the square, and is formally stated as:

The Quadratic Formula: For ax2 + bx + c = 0, the values of x which are the solutions of the equation are given by:

$x = \dfrac{-b \pm\sqrt{b^2 - 4ac\,}}{2a}$

For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root. And it's a "2a" under there, not just a plain "2". Make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back in" on your test, and you'll mess yourself up. Remember that "b2" means "the square of ALL of b, including its sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive.

In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!

Here are some examples of how the Quadratic Formula works:

• Solve x2 + 3x – 4 = 0

This quadratic happens to factor:

x2 + 3x – 4 = (x + 4)(x – 1) = 0

...so I already know that the solutions are x = –4 and x = 1. How would my solution look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, my solution looks like this:

$x = \dfrac{-(3) \pm \sqrt{(3)^2 - 4(1)(-4)\,}}{2(1)}$

$= \dfrac{-3 \pm \sqrt{9 + 16\,}}{2} = \dfrac{-3 \pm \sqrt{25\,}}{2}$

$= \dfrac{-3 \pm 5}{2} = \dfrac{-3 - 5}{2},\, \dfrac{-3 + 5}{2}$

$= \dfrac{-8}{2},\, \dfrac{2}{2} = -4,\, 1$

Then, as expected, the solution is x = –4, x = 1.

http://www.purplemath.com/modules/quadform.htm

Algebra 2

End Behavior of a Function

The end behavior of a polynomial function is the behavior of the graph of f(x)$f\left(x\right)$ as x$x$ approaches positive infinity or negative infinity.

The degree and the leading coefficient of a polynomial function determine the end behavior of the graph.

The leading coefficient is significant compared to the other coefficients in the function for the very large or very small numbers. So, the sign of the leading coefficient is sufficient to predict the end behavior of the function.

Degree	Leading Coefficient	End behavior of the function	Graph of the function
Even	Positive	f(x)→+∞, as x→−∞f(x)→+∞, as x→+∞$\begin{array}{l}f\left(x\right)\to +\infty ,\text{ as }x\to -\infty \\ f\left(x\right)\to +\infty ,\text{ as }x\to +\infty \end{array}$	Example: f(x)=x2$f\left(x\right)=x^2$
Even	Negative	f(x)→−∞, as x→−∞f(x)→−∞, as x→+∞$\begin{array}{l}f\left(x\right)\to -\infty ,\text{ as }x\to -\infty \\ f\left(x\right)\to -\infty ,\text{ as }x\to +\infty \end{array}$	Example: f(x)=−x2$f\left(x\right)=-x^2$
Odd	Positive	f(x)→−∞, as x→−∞f(x)→+∞, as x→+∞$\begin{array}{l}f\left(x\right)\to -\infty ,\text{ as }x\to -\infty \\ f\left(x\right)\to +\infty ,\text{ as }x\to +\infty \end{array}$	Example: f(x)=x3$f\left(x\right)=x^3$
Odd	Negative	f(x)→+∞, as x→−∞f(x)→−∞, as x→+∞$\begin{array}{l}f\left(x\right)\to +\infty ,\text{ as }x\to -\infty \\ f\left(x\right)\to -\infty ,\text{ as }x\to +\infty \end{array}$	Example: f(x)=−x3$f\left(x\right)=-x^3$

To predict the end-behavior of a polynomial function, first check whether the function is odd-degree or even-degree function and whether the leading coefficient is positive or negative.

Example:

Find the end behavior of the function x4−4x3+3x+25$x^4-4x^3+3x+25$.

The degree of the function is even and the leading coefficient is positive. So, the end behavior is:

f(x)→+∞, as x→−∞f(x)→+∞, as x→+∞$\begin{array}{l}f\left(x\right)\to +\infty ,\text{ as }x\to -\infty \\ f\left(x\right)\to +\infty ,\text{ as }x\to +\infty \end{array}$

The graph looks as follows:

 https://www.varsitytutors.com/hotmath/hotmath_help/topics/end-behavior-of-a-function