GRAPHING
QUADRATIC FUNCTIONS
Objectives:
Recall 1. In the past we put quadratic equations in the form
y = ax2 + bx + c.
This makes the equation look nice and become more readable. For the purposes of graphing, we would rather alter this form slightly. Recall that we represented lines in standard form as
y = mx + b.
Then we were able to read the critical information about slope and y-intercept directly from the form of the equation. The slope is m and the y-intercept is b. We wish to do the same with the graphs of quadratic functions. We will use several ideas that were touched on in past hours to provide a scheme for knowing how to graph a quadratic function.
Recall 2 (The standard parabola centered at the origin.)
If you sketch the graph of the quadratic equation y = x*x, you find a curve that looks like the curve below This curve is called a parabola. The first thing to notice about a parabola is its symmetry. If you flip this parabola about the y-axis, you get the exact same parabola. Every parabola has such a line of symmetry. In the case of the parabola y = x2, the y-axis is the line of symmetry. The point at which the parabola meets its line of symmetry is called the vertex of the parabola.
Now, the basic shape and orientation will be determined
by the coefficients of the quadratic y = ax2 + bx + c.
We take one possibility at a time. Suppose that the coefficients b and c are both 0. In this case our parabola is given by the equation
y = ax*x.
The question should now be: What role does
the coefficient
a play in orienting or shaping the graph? To
find
out, let a be a
few different values and compare the values of a with the
corresponding
shapes of the sketches that we get. The sketches below
suggest the
role of a.
Note
If the coefficient is negative, the parabola faces
downward.
If the coefficient is positive, then
the parabola faces
upward.
The absolute value of
the coefficient determines the
sharpness of the curve at the vertex--the
higher the absolute value, the
steeper and sharper the
parabola.
Recall 3 . (The general case y = ax2 + bx + c.)
If we can put the equation into the form
y = a(x - h)2 + k,
where
h and k
are specific real numbers then we
would know the orientation
and position of
the parabola that is the graph of the
quadratic equation. It turns
out that this parabola has a
vertex at the point (h,k).
Again, the steepness of the
parabola, or the sharpness at the vertex is
determined by the
coefficient a--the larger its absolute
value, the steeper the
parabola.
For example, the curve
y = 2(x - 5)2 + 7
is a parabola opening upward with vertex at (5,7). The line of symmetry in this case is the vertical line x = 5.
Recall 4. (Changing y = ax2 + bx + c to the form y = a(x - h)2 + k)
Any quadratic
equation of
the form y = ax2 + bx
+ c can be changed to the form y = a(x -
h)2 + k,
if the proper values
of h and k can be found. The
process of finding such h and
k is called completing the
square.
Here's how it works. Start with y = ax2 + bx
+ c and take a factor of a out from the first two terms of the right-hand side.
(Remember a is different from 0, otherwise we would have the linear equation
y = bx + c. ) We get
Notice that this is in the form y = a(x - h)2 + k, with h =-(b/2a) and k=(c-a(b/2a)2)
This means that our parabola has vertex at (-(b/2a), (c-a(b/2a)2))
Example 1. Find the axis of symmetry and the vertex of the graph of the function
f(x) = x2 - 6x + 11.
Solution: List the coefficients a = 1, b = -6, and c = 11. By completing the square for the equation y = x2 - 6x + 11, we find
y = (x2 - 6x ) + 11
=(x2-6x+9) + 11 -9
=(x-3)2 + 11 - 9
= (x - 3)2 + 2.
Therefore, the parabola faces upward (a is positive) and it has a vertex at (3,2) with a vertical line of symmetry x = 3.
Example 2. Find the axis of symmetry and the vertex of the graph of the function
f(x) = -2x2 + 16x - 35.
Solution: List the coefficients a = -2, b = 16, and c = -35. By completing the square for the equation y = -2x2 + 16x - 35, we find
y = -2(x2 - 8x) - 35
= -2(x2-8x+16)- 35 + 2*16
= -2(x-4)2 - 35 + 32
= -2(x-4)2- 3.
Therefore, the parabola faces downward (a is negative) and it has a vertex at (4,-3) with a vertical line of symmetry x = 4. Notice that the parabola has a sharper steepness than the one in the previous example, because a = 2.
A quadratic function is in the form f(x) = ax2 + bx + c. A quadratic function graphs into a parabola, a curve shape like one of the McDonald's arches. Every quadratic function has a vertex and a vertical axis of symmetry. This means it is the same graph on either side of a vertical line. The axis of symmetry goes through the vertex point. The vertex point is either the maximum or minimum point for the parabola.
| a>0
a<0 | graph opens up and the
vertex is a minimum
graph opems down and the vertex is a maximum |
| b2 - 4ac > 0
b2 - 4ac = 0 b2 - 4ac < 0 | Graph
has 2
x-intercepts
Graph has 1 x-intercept Graph has no x-intercepts |
| Vertex Point | at x = -b/2a
to find y substitute above value for x in the expression of the function |
| x-intercepts | (-b+sqrt(b2-4ac))/2a and
(-b-sqrt(b2-4ac))/2a if
b2-4ac>=0 none if b2-4ac<0 |
| y-intercept | c value in ax2 + bx + c |
EXERCISES
1. Plot the
points (x,f(x)) at x = -4, -3, -2,
-1, 0, 1,
2, 3, and 4 for the function f(x) =
x2 -6x + 2.