A few comments on Quadratic Functions on SL IB exams…
- Questions focusing on quadratics seem to be mostly on Paper 1
- You can not use your calculator on Paper 1
- So learn on to do things with out your calculator too!
Definition of a Quadratic Function
A quadratic function can be written in the form of $ax^2+bx+c$ where $a \neq 0$. If $a = 0$ then the function would be linear and not quadratic!
Terminology
Y-Intercepts are the points on the graph where $x=0$. That is the points where the graph crosses the y-axis. To find the y-intercepts you simply substitute 0 in for x and simplify. If the function is given in the form of $ax^2+bx+c$ then the y-intercept is simply the value of c.
X-Intercepts are the points on the graph where $y=0$. Or in other words the points where the graph crosses the x-axis. Finding x-intercepts is a bit trickier than for the y-intercepts. There are several methods to do so.
Roots and Zeros are the values of the independent variable (often x) that make the entire function zero. These coincide with the x-intercepts. The only difference being that the x-intercepts are coordinate points on a graph, roots or zeros are a value of a variable.
Standard Form or Vertex Form this form goes by different names, but the form I am referencing is $a(x-h)^2+k$. This form can be challenging to get but the form tells us a great deal more information. The vertex of the parabola is at $(h,k)$.
Axis of Symmetry this is the imaginary line that "splits" the parabola vertically into two symmetric pieces. This line naturally goes through the vertex so the axis of symmetry is given by the line $x=h$ where h is the value from the standard form. With a little bit of magic and algebra it can be shown that the axis of symmetry occurs at:
(1)Getting into Standard Form
The IB likes to ask questions that require you to go from general form $ax^2+bx+c$ to standard form $a(x-h)^2+k$. There are a few ways to do this, but do remember you need a non-GDC method in your back pocket.
GDC can be used to find the minimum or maximum of a function. In the case of a quadratic function the minimum or maximum also occur at the vertex. It only requires a few buttons pushes to get it done.
If you don't know how you can look here. You will need to scroll down to find the description of how to do it.
Completing the Square is one of the most feared and dreaded techniques in an advanced algebra course. Luckily, I don't think you'll need to do this for your IB exam. You can see all the details on my page on completing the square. Summary looks something like this:
(2)where
(3)and
(4)This also results in the fact that the vertex of a parabola is at $(x,y)=(h,k)=\left ( \frac{-b}{2a},c - \frac{-b^2}{4a} \right )$.
Axis of Symmetry
Since the axis of symmetry must pass through the vertex then the equation for the axis of symmetry must be:
(5)This formula is on the IB formula handbook! Which means you can simply write down the x-coordinate of the vertex if you are given a quadratic equation in the form of $ax^2+bx+c$. Once you have the x-coordinate you can simply substitute that value back into the function and then get the y-value. If you can understand and remember the last 2-3 sentences you should be golden on at least 50% of IB quadratic questions.
Example
Write the function $2x^2+5x+3$ in the form $a(x-h)^2+k$.
First we find the axis of symmetry at $x= \frac{-b}{2a}=\frac{-5}{4}$. This value then goes back into the function $2(\frac{-5}{4})^2+5(\frac{-5}{4})+3 = -0.125$.
Which puts the vertex at $(x,y)=(-1.2 , -0.125)$. And thus standard form $2(x-1.2)^2-0.125$
Finding Zeros (Roots)
Quadratic Formula this formula is on your IB Formula Booklet. I'll put it here for reference too:
(6)This is a pretty simply "plug-and-chug" equation. The letters in the formula correspond to the general quadratic form $ax^2+bx+c$. Its key to remember that in general two solution come from the Quadratic Formula… this is due to the $\pm$ in the equation.
The number of zeros and the type of zeros can be determined from the discrimnate:
(7)This is nothing more than the part of the quadratic equation that is in the square root. If the discriminate is zero the there will be only one solution. There are still two zeros, but they are repeated. This also coincides with the vertex being on the x-axis! If the discriminate is greater than zero there will be two solutions and they will both be real. If the discriminate is negative there will be two solution but both zeros with be imaginary.
| Discriminate | Solutions & Zeros |
| $\Delta = 0$ | 1 solution, two repeated real zeros |
| $\Delta > 0$ | 2 solutions, two distinct real zeros |
| $\Delta < 0$ | 2 solutions, two distinct imaginary zeros |
GDC - Graphing Display Calculator
Your calculator has a function for finding the zero of any function. This can be found under the "calc" option at the top right.
Look here for screen shots of exactly how to do it. You will need to scroll down a bit to find the part on zeros.
Factoring
This third option is a good option if you are "good" at factoring and can do it quickly. Otherwise the above two methods are more "sure-fire." Factoring can be a waste of time as not all quadratics can be factored, at least not with real numbers…
Factoring for those unfamiliar with it looks like:
(8)Where $d+e=b$ and $(d)(e)=c$.
For those who really like to factor or are faced with a question that requires them to factor see the topic below regarding the Factor Theorem.
Factor Theorem (Plus the Fundamental Theorem of Algebra)
These two theorems put together essentially say that every polynomial has at least one zero, but more specifically a polynomial has as many zeros as its largest exponent. That is a 5th order polynomial has 5 zeros. Note not all zeros have to be real and they can be repeated in pairs.
An nth order polynomial can be written in the form of:
(9)Where $z_n$ is the nth zero and a is simply a coefficient. The usefulness of this comes from the fact that you can write the equation for any polynomial if you know the zeros and one other point. The additional point is needed to find the value of the leading coefficient (a). Also knowing the zeros of a polynomial allows you to write the polynomial in factored form.
Example
The graph below shows a third order polynomial. We can be pretty sure it's 3rd order as it has three zeros. It is possible that it's a higher order polynomial, but that would just be mean…
From the graph we can see that the zeros occur at $(-2,0),(1,0),(3,0)$. From this we can write the equation for the polynomial as:
(10)We still need an additional point to find the value of a. Looking at the graph we can see that the function goes through the point $(0,3)$. Substituting this point into the equation we get:
(11)Solving for a we find that $a=\frac{1}{2}$ producing the final equation of:
(12)
Post preview:
Close preview