The shape of a parabola is like an arch. Parabolas are modeled by quadratic functions of the form y = ax2 + bx + c. The first term cannot equal zero because it will make the function linear, or a straight line. Therefore, the coefficient a cannot be zero.
Graph each quadratic function by making a table of values.
1.
y = x2 + 1
Alternative Solutions:
First, choose integer values for x. Evaluate the function for each x-value. Graph the points and connect them with a smooth curve.
2.
y = –x2
Alternative Solutions:
In Example 1, the lowest point,
or minimum, of
the graph of y = x2 + 1 is at (0, 1). Since the
coefficient of x2 is positive, the graph opens upward.
In Example 2, the highest point, or maximum, of the graph of y
= –x2 is at (0, 0). Since the coefficient of x2
is negative, the graph opens downward.
As with linear functions, in a
quadratic function x is the independent variable and y is the
dependent variable. Since the graph of a quadratic function extends forever to
the left and to the right, the domain (x values) of a quadratic function
is the set of all real numbers. For a quadratic function with a graph that
opens upward, the range (y values) is all real numbers greater than or
equal to the minimum value. For a quadratic function with a graph that opens
downward, the range is all real numbers less than or equal to the maximum
value.
The maximum or minimum point of a
parabola is called the vertex.
The vertical line containing the vertex of a parabola is called the axis of symmetry. If you
fold the graph of y = x2 + 1 or y = –x2
along the axis of symmetry, the two halves of each graph will coincide. In both
examples, the axis of symmetry is the line x = 0.
You can use the rule below to find
the equation of the axis of symmetry.
Example
3.
Use characteristics of quadratic functions to graph y = x2
– 4x – 1.
A.
Find the equation of the axis of symmetry.
B.
Find the coordinates of the vertex of the parabola.
C.
Graph the function.
Alternative Solutions:
A. First identify a, b, and c.
Now, find the equation of the axis of symmetry.
B. Next, find the vertex. Since the equation of the axis of symmetry is x = 2, the x-coordinate of the vertex must be 2. Substitute 2 for x in the equation y = x2 – 4x – 1 to solve for y.
C. Construct a table. Choose some values for x that are less than 2 and some that are greater than 2. This ensures that points on each side of the axis of symmetry are graphed.
Models of quadratic functions can be
found in the real world.
Architecture
Link
4. The Exchange House in London, England, is supported by a steel arch shaped
like a parabola. This parabola can be modeled by the quadratic function y =
– 0.025x2 + 2x, where y
represents the height of the arch and x represents the horizontal distance from one end of the base in meters. What
is the highest point of the parabolic arch?
Alternative Solutions:
The
highest point of the arch is the y-coordinate of the vertex. Find the equation
of the axis of symmetry for h(x) = – 0.025x2
+ 2x.
Next,
find the vertex. Since the equation of the axis of symmetry is x = 40, the
x-coordinate of the vertex must be 40. Substitute 40 for x in the
function h(x) = – 0.025x2 + 2x.
Then solve for h.
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