Lesson 4
1. Lesson 4
1.7. Explore 2
Module 3: Quadratics
Self-Check 2
Convert the vertex form of Jean-Guy’s jump, y = −0.2(x − 3)2 + 1.8, to the standard form by showing each step. Answer
Did You Know?
Although the vertex form of the quadratic function for Chris’s jump is written as y = −0.3(x − 2)2 + 1.2, it can more accurately represent the situation if it were written using h for height and d for distance. It would, of course, still be a quadratic function, but it would look like h = −0.3(d − 2)2 + 1.2. If this form is used, the vertical axis on the graph would need to be labelled as h and the horizontal axis as d.
You are given the vertex of a parabola and the intercepts and asked to write the function in vertex form. The h and k values are easily written from the vertex, but how do you get the value for a?
Try This 2
For the following graph, the vertex is at (−2.5, −4.9). From this information, the function can be written as y = a(x + 2.5)2 − 4.9.
- Substitute the x- and y-values from one of the intercepts into the function and solve for a. What value for a do you get?
- Verify by substituting the x-value from the second intercept and the a-value you just calculated into the function y = 0.4(x + 2.5)2 - 4.9. Did you get the y-value of the second intercept—in this case, y = 0?
Read “Example 2: Determining the equation of a parabola using its graph” on page 357 of the textbook. As you read the example, consider how two points can be enough to determine the quadratic equation if one of the points is the vertex.
Read “Example 4: Solving a problem that can be modelled by a quadratic function” on pages 359 to 361 of the textbook. As you read, think about how determining the height of the ball in part c) is similar to determining a in part a). How is it different?