Equivalent Quadratic Forms. Lesson 7.2. In Lesson 7.1, you were introduced to polynomial functions, including 2nd-degree polynomial functions, or quadratic functions. The general form of a quadratic function is y=ax 2 +bx+c.
In Lesson 7.1, you were introduced to polynomial functions, including 2nd-degree polynomial functions, or quadratic functions.
Recall from Chapter 4 that every quadratic function can be considered as a transformation of the graph of the parent function y =x2 .
identifies the location of the vertex, (h, k), and the horizontal and vertical scale factors, a and b.
To rewrite the equation with a single scale factor, first move the denominator out of the parentheses, which gives the equation
This new form, , is called the vertex form of a quadratic function because it identifies the vertex, (h, k), and a single vertical scale factor, a.
The y-coordinate of any point along the x-axis is 0, so the y-coordinate is 0 at each x-intercept.
You should find that the x-intercepts of the graph of y =1.4(x-5.6)(x+3.1).
are 5.6 and -3.1.
The vertex form is y =(x-1)2-4
The x-intercepts are 1 and 3, so the factored form is y= a(x+1)(x-3). To verify that the scale factor, you can solve for it analytically. Substitute the coordinates of another point on the parabola such as (2, 3) into the equation for x and y.
Prop up one end of the table slightly. Place the motion sensor at the low end of the table and aim it toward the high end.
With tape or chalk, mark a starting line 0.5 m from the sensor on the table.
Set up your calculator to collect data for 6 seconds. When the sensor begins, roll the can up the table.
The shape of the graph is parabolic. A quadratic function would model these data. The second differences, D2, are almost constant around 0.06, which implies that a quadratic
model is appropriate.
Mark the vertex and another point on your graph. Approximate the coordinates of these points and use them to write the equation of a quadratic model in vertex form.
vertex: (3.2, 4.757);
point: (5.2, 1.397);
Horizontal stretch: 5.2-3.2 or 2
Vertical stretch: 1.397 -4.757or -3.36
From your data, find the distance of the can at 1, 3, and 5 seconds. Use these three data points to find a quadratic model in general form.
(1, 1.046), (3, 4.756), and (5, 1.993);
1.046=a(1)2 + b(1) +c
4.756=a(3)2 + b(3) +c
1.993=a(5)2 + b(5) +c
a= -0.81, b= 5.09 and c = -3.24
Y= -0.81x2 +5.09x -3.24
Mark the x-intercepts on your graph. Approximate the values of these x-intercepts. Use the zeros and the value of a from Step 5 to find a quadratic model in factored form.
Zeros: (0.7, 0) and (5.6, 0);
Verify by graphing that the three equations in the last three steps are equivalent, or nearly so. Write a few sentences explaining when you would use each of the three forms to find a quadratic model to fit parabolic data.
y= -0.81x2 +5.09x -3.24