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## Solving a Quadratic Equation
• Students are able to recognize formulas for quadratic functions in standard
vertex • Given a graph of a quadratic function, f(x), that shows the locations of
two distinct x-intercepts and the • Given a graph of a quadratic function, f(x), that shows the coordinates of
the vertex and one other point • Given an explicit formula for a quadratic function, students are able to
determine the x- and y-coordinates • Students should be able to determine the maximum/minimum of a quadratic
function by converting the
Most students are already very familiar with quadratic functions, standard
form and factoring, however,
students are: • Based on the information that you can see on the graph,
which way of writing a quadratic function • How do the points you can see on the graph help you to
figure out the numbers that need to go into the • What added information do we gain from the extra point (i.e. not the zeros or vertex) on the graph? • How can you know the sign of the leading coefficient just by looking at the graph? Generally, students understand how to use zeros and/or
vertex points in equations, but oftten have a
Note: The discussion of completing the square in the
Chapter 5 Tools section (page 239) of the textbook
The most common mistakes are (i) not factoring a out of
everything in Step 1 and (ii) not distributing a From the vertex form now found for h(x), students should
be able to easily comment about how the
do since the students must factor out a negative number.
The current price of a ticket at a local movie theater is
$8. At this price, the theater sells 1200 tickets You can help the students tackle this problem by putting
the following subquestions to direct them. If the students finish this quickly, you could help them
get started on their homework problems Wrap up the class by pointing out that the vertex form of
a quadratic function is very useful for optimization of
If you consider a quadratic function that is expressed in vertex form, for example: you might wonder just why it is that the vertex appears at
the point (2, 4). We will explain this by using To refresh your recollection of how transformations are
carried out algebraically and graphically, Table 1 shows To convert the basic quadratic y = x^2 into the new
quadratic , you need to perform three 1. Horizontal translation (in this case by 2 units to the right) 2. Vertical stretch/reflection (in this case there was a
vertical compression to half the usual height, 3. Vertical translation (in this case by 4 units upwards). So, the graph of will
be a wide, upside down parabola that is displaced to the left and Figure 3: Note that the vertex of the basic quadratic y = x^2 has been shifted from the point (0, 0) to the point (2, 4). In a more general setting, in order to transform the basic
quadratic equation y = x^2 into the usual formula for you would have to carry out the following transformations: 1. A horizontal shift of h units to the right. The horizontal shift will move the vertex of the quadratic
from (0, 0) to (h, 0) and the vertical shift will move occurs at the point (h, k). |