<p>What is a quadratic equation? It is this: f(x)= ax^2 + bx +c, where a, b and c are constants. In other words, a quadratic equation is a univariate (has one variable-x) polynomial equation of the second degree. It is also called a parabola.</p>
<p>What does it look like?
If a is positive, it is basically a big U. If a is negative, it is basically a U that is open downward (google this if you don’t understand me)</p>
<p>Transformations:
- You can move it up and down by changing the value of c. For example, if you had f(x)= 2x^2 +5 and you changed it to f(x)= ax^2 +2, you moved the graph down by 3 units.
- You can move it left and right by …just take this as an example. If f(x)= 2x^2 and g(x)=2(x-2)^2, g(x) is basically f(x) moved two units to the right. Basically, you have to change the number that is in the parentheses with x- adding moves the graph to the left, and subtracting moves it to the right (the opposite of moving the graph vertically).
- As the “a” in f(x)= ax^2 +c gets bigger, the U becomes more narrow- and vice versa. Why? Basically, the function reaches higher (or lower) values faster when “a” becomes bigger.</p>
<p>What is a “vertex”?
It is the lowest point for the graph if a in the function is positive, and the highest point if a is negative (google some pictures so you can have visual aid). The vertex is special because it signifies a change in the sign of the slope. In layman’s terms, if the function was always going up, it will start going down- and vice versa. Also, points that are equally far from the vertex will always have the same y value. For example, if you vertex is (3,1), and you have (5,3) and (1,y), y would equal 3. To find the slope the x coordinate of the slope, use this formula: x=(-b/2a). The formula is basically like taking the derivative and equaling it to zero (when you take calculus you’ll get this). Here is a neat trick: let’s say you have (2,5) and (4,5) as points on your graph. Obviously, form what I said before, these two are equally far from the vertex; therefore the x value of the vertex is the mean of 2 and 4: 3! One final thing, if you have (-1,5), (1,4) and (x,5), x will be 3. Can you tell me why?</p>
<p>The infamous quadratic equation and its implications:
The quadratic equation finds out x value when a quadratic function equals zero (x-intercept). The formula is… just google it. The b^2 -4ac is the determinate (you know, whatever is under the root). If the determinate is positive, the function will equal zero at two different values of x. This also means that the function will touch the x axis twice and change in sign during the interval between those 2 x values. To keep it simple, let’s say you got x^2 -5x-6. Firstly, you know that the graph is basically a U. Now you use the quadratic equation and you find out the you have two values for 2 and 3. So the function will start really high up (positive) and keep going down till it hits the x axis at 2. Then it will go into the negatives. It will start rising up and hit the x axis again at 3, and continue rising. In other words, if the determinate is positive, the function will cross the x axis. </p>
<p>F(x) sign: +++++++0-------0++++++++
Corresponding value of x: 0 1 2 2 2.5 3 4 5 6 </p>
<p>If the determinate is zero, the function will touch the x axis but not cross it. So if a is positive the graph will always be above the x axis. If a is negative, it will always be below. By the way, the solution to the quadratic equation in this case is the vertex. </p>
<p>If the determinate is negative, the graph will never touch the x axis, so the graph will either be always above or always below the axis, depending on the sign of a. The solution to the quadratic equation will be two complex numbers ( will matter if you take an SATII in math). The normal SAT doesn’t deal with this case.</p>
<p>Factoring: Using the quadratic equation is cumbersome. Instead of wasting our time we will just factorize the function. Look at this:
f(x)= x^2 -5x +6 = (x-6)(x+1) What we did here is ask ourselves this: what 2 numbers do I multiply to get -6 (c)and add to get -5(b)-- {-1, 6} of course. So we will just put them into two separate sets of parentheses and add in x’s. Those 2 x values form the solution set for the quadratic equation and are the x values when y equals zero (x-intercept). </p>
<p>By the way, if you have something like x^2+ 5x (no c), just take x and whatever is “a” as the factor. So, if you have 2x^2 -15 x, you’ll factorize it like this.
f(x)= 2x^2 - 15x= 2x(x-7.5). Your solution set is {0, 7.5}. </p>
<p>If you got f(x)= a(x-d)^2+ c (no x to the power 1), your solution set is only d, which is also the x value for the vertex. By the way, in this case, the determinate is zero and the function never crosses the x-axis.</p>
<p>I’ll post a bunch of questions later. If anyone has anything to add, feel free to do so.</p>