Quadratic Forms

Quadratics are written in three basic forms. They each have their own uses. The names for the three forms are unimportant and may depend on the textbook, teacher or country. The IB does not give names for the different forms. The names I choose I like and make sense to me…

Standard/General Form

This is the form that most students are comfortable with, but is one of the least useful forms.

(1)
\begin{equation} y=ax^2+bx+c \end{equation}

By least useful I mean that it is challenging to make an accurate sketch of the function from this form. The c value represents the y-intercept and a represents a vertical dilation neither of which allow for easy graphing. Plus b does "strange" things.

This form is handy because the parameters (a,b,c) correspond to the parameters in the quadratic formula:

(2)
\begin{align} x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \end{align}

It is also very common to make use of the formula for the axis of symmetry in IB problems which also makes use of the parameters in the standard form:

(3)
\begin{align} x=\frac {-b}{2a} \end{align}

Factored Form

This form appears like:

(4)
\begin{equation} y=a(x-p)(x-r) \end{equation}

Where p and r are the zeros of the function. Graphically the zeros are where the function crosses the x-axis. The sign of a tells you which direction the parabola opens. Knowing the points that the graph crosses the x-axis and which way the parabola opens allows a quick and moderately accurate sketch of the function.

Vertex Form

This form appears like:

(5)
\begin{equation} y=a(x-h)^2+k \end{equation}

In this form h represents a horizontal shift and k represents a vertical shift. If both h and k are zero then the vertex of the parabola would be at $(0,0)$. There for the coordinates of $(h,k)$ give us the vertex of the parabola! The sign of a tells you which direction the parabola opens

A Comment on All Forms

In all the forms a plays the same role. In fact if you move from one form to another a will always have the same value, which is a very handy thing to remember.

Converting Between Forms

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Getting to Vertex Form

Getting into Vertex Form often seems to be the trickiest bit for students. Below are explanations how to do it with and without your GDC.

Without GDC

The IB likes to ask questions that require you to go from standard form $ax^2+bx+c$ to vertex 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. An "easy" method is to use the so-called axis of symmetry.

The axis of symmetry is an imaginary line vertical that splits the parabola into two equal parts. The equation for this line is given by:

(6)
\begin{align} x=\frac{-b}{2a} \end{align}

This formula is on the IB formula handbook! The axis of symmetry must pass through vertex of the parabola, so the equation above also gives us the x-coordinate of the vertex. 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.

With GDC

A GDC can be used to find the minimum or maximum of a function. In the case of a quadratic function the minimum or maximum is 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.

Example - Without GDC

Write the function 2ドルx^2+5x+3$ in the form $a(x-h)^2+k$.

First we find the axis of symmetry at

(7)
\begin{align} x= \frac{-b}{2a}=\frac{-5}{4} \end{align}

This value then goes back into the function

(8)
\begin{align} 2(\frac{-5}{4})^2+5(\frac{-5}{4})+3 = -0.125 \end{align}

Which puts the vertex at:

(9)
\begin{equation} (x,y)=(-1.2 , -0.125) \end{equation}

And thus vertex form

(10)
\begin{equation} y=2(x-1.2)^2-0.125 \end{equation}

Want to add to or make a comment on these notes? Do it below.

John (guest) 12 May 2014 17:18

For the example without GDC for getting to vertex form, shouldn't the equation be {2(x+1.2)^2}-0.125 if the x value of the vertex is -1.2?

by John (guest), 12 May 2014 17:18
carlota (guest) 02 Jan 2018 16:42

Yes, if x is equivalent to h and x = -1.2

then wouldn't it be y = 2(x+1.2)2 - 0.125

as the two negatives make it a positive?

by carlota (guest), 02 Jan 2018 16:42

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