If I am given the equation 2x + 4y = 8, what are
the names of the two easily found points on its
graph?
If I am given the equation 2x + 4y = 8, what are
the names of the two easily found points on its
graph?
___________________ and ____________________
___________________ and ____________________
Find them for this equation.
Find them for this equation.
Graph the equation.
Graph the equation.
What if I am given the equation y = x2 – 2x – 3?
Can I find those same points for its graph?
What if I am given the equation y = x2 – 2x – 3?
Can I find those same points for its graph?
Find them for this equation.
Find them for this equation.
We have spent time solving quadratic equations
by factoring. There are 3 different names we can
use for the answers:
 solution
 root – The founder
of Algebra, Arabic
mathematician alKhwarismi, saw the
variable as the root
out of which an
equation grows.
 zero – When the equation is equal to zero,
these are the answers. Also, when you
substitute in the solutions, the result is zero.
These three terms are interchangeable.
When we graphed linear equations, we could
always find the ______________ and the
_______________. We can still find them.
 We find the ___________ by substituting in ___
for ___ and solving for ___.
 We find the ___________ by substituting in ___
for ___ and solving for ___.
How many x-intercepts do you think you could
have for a quadratic equation? Why?
How many y-intercepts do you think you could
have for a quadratic equation? Why?
1)
a)
Find the zeros for each quadratic function.
f(x) = x2 – 2x – 8
b)
f(x) = x2 + 8x + 16
c)
f(x) = -x2 + 7x – 10
d)
f(x) = -3x2 + 2x + 5
When you are finding the zeros of a function, you
are also finding the x-intercepts of the function.
(Remember that all x-intercepts have a y-value of
zero.)
It is possible to not have an x-intercept of a
function. For now, we will give problems with at
least one x-intercept.
2)
What are the x-intercept(s) of each
quadratic function?
a)
f(x) = x2 – 2x – 8
b)
f(x) = x2 + 8x + 16
c)
f(x) = -x2 + 7x – 10
d)
f(x) = -3x2 + 2x + 5
Another easy point to find is the y-intercept.
3)
Find the y-intercept for each quadratic
function.
a)
f(x) = x2 – 2x – 8
b)
f(x) = x2 + 8x + 16
c)
f(x) = -x2 + 7x – 10
d)
f(x) = x2 + 5x + 6
f(x) = -x2 + 5x + 6
f(x) = -3x2 + 2x + 5
Some parabolas open upward, others open
downward. What is the difference between the
equations whose graphs open upward versus
those that open downward?
The graph of a quadratic function is a curve
called a parabola. A parabola is a pointy ushape.
f(x) = x2
f(x) = -x2
If the equation is in the form f(x) = ax2 + bx + c, the
value of a determines which direction the
parabola opens.
If a is __________ (______), the parabola opens
___________.
If a is __________ (______), the parabola opens
___________.
4)
a)
Which way does the parabola open?
f(x) = x2 – 2x – 8
b)
f(x) = x2 + 8x + 16
c)
f(x) = -x2 + 7x – 10
d)
f(x) = -3x2 + 2x + 5
So far, we know:
 x-intercept(s)
 y-intercept
 which way the parabola opens
In this learning target, all we are asked to do is
sketch the graph. That means it does not need to
be 100% precise. We should, however, try to make
it look as reasonable as possible.
Now, plot the points you know. (Question #5)
We can use the idea of symmetry to help us fill in
a point and figure out about where the vertex is.
It’s kind of hard to figure out the exact parabola
with just 2 or 3 points. How else do you think we
can fill in the shape?
Here are a couple more things about parabolas
that may be helpful to know:
 Every parabola has a highest or lowest point
called the _________.
 Every parabola has _____________. In this
case, that means we can “fold” the
parabola on an imaginary vertical line and it
will lie on top of itself.
If the graph has two x-intercepts, where do you
think the fold line we talked about would fall?
If the graph has one x-intercept, where do you
think the fold line we talked about would fall?
Go back to where you graphed the points you
found. Draw in with a dashed line where you
believe the fold line would be. (Question #6)
Let’s see if we can find one last point. Use the fold
line on your graph and the y-intercept to find the
mirror image of the y-intercept if the mirror is the
fold line. Mark this point on each of your graphs.
(Question #7)
Hopefully now you have 4 points (or 3 points in the
case of having one x-intercept). Now sketch in
the parabola shape for each graph. (Question
#8)
5)
Plot the points you know.
6)
Draw in a dashed line to represent the fold
line.
7)
Graph the point that is the mirror image of
the y-intercept.
8)
Sketch the parabola.
a)
b)
c)
f(x) = -x2 + 7x – 10
d)
f(x) = -3x2 + 2x + 5
f(x) = x2 – 2x – 8
f(x) = x2 + 8x + 16