In the Standard Form of a Quadratic Relationship, by just looking at the expression, we can determine
* the y-intercept
* whether the parabola is concave up (cup-shaped) or concave down (umbrella-shaped)
In the Factored Form of a Quadratic Relationship, by just looking at the expression, we can determnine
* the zeros, or x-intercepts, of the expression
* whether the parabola is concave up or concave down
We are now going to look at the Vertex Form of a Quadratic Expression.
Like the Standard and Factored Forms, the Vertex Form of a Quadratic Relation also has a few characteristics:
* the vertex can easily be found
* the concavity can easily be determined
The Vertex Form of a parabola is in the form y = a(x-h)² + k
When an expression is in Vertex Form, the vertex is (h,k). Also, we must take the opposite sign of h.
If the vertex and another point on the parabola is known, it is easy to determine the equation of the parabola:
Substitute the coordinates of the vertex into h and k, and substitute in the coordinates of the other point for x and y, and then solve for a, remembering to use Order of Operations (BEDMAS).
Recall that if a > 0, then the parabola opens upward. If a < 0, then the parabola opens downward.
The equation of a parabola can be turned into Standard form by expanding and collecting like terms (this will involve FOIL, and Order of Operations)
Find the vertex, the axis of symmetry, and the direction of opening for the following quadratic relations.
(a) y = 2 (x - 5)² + 6
The coordinates of the vertex are (5,6); the axis of symmetry is x = 5; and the vertex opens up (positive a value)
(b) y = - 1/2 (x + 7) - 4
The coordinates of the vertex are (-7,-4); the axis of symmetry is x = -7; and the vertex opens down.
A stick is thrown into the air. Its height H in meters after t seconds is
H = - 2 (t - 3)² + 20.
(a) In what direction does the parabola open? How do you know?
(b) What are the coordinates of the vertex, and what does this represent?
(c) From what height was the stick thrown?
(a) The parabola opens downward, because a = -2. Also, it makes sense that if a stick is thrown, it will first go up, and then gravity will bring it back down again - thus, it makes the shape of a concave down parabola.
(b) The coordinates of the vertex are (3,20). This means that the stick reached its maximum height of 20 meters after 3 seconds.
(c) The stick was thrown when t = 0. Substitute this value in for t, and solve for H.
H = -2 (0 - 3)² + 20
H = -2 (3)² + 20
H = -2 (9) + 20
H = -18 + 20
H = 2
Therefore, the stick was thrown from a height of 2 m.
A parabola passes through the point (4,40), and has a vertex (-3,-9). FInd its equation in vertex form.
Since the vertex and a point are given, we can use vertex form.
y = a(x + 3)² - 9
Also, we know a point on the line.
40 = a(4 + 3)² - 9
40 = a(7)² - 9
40 = a (49) - 9
49 = 49a
a = 1
Therefore, the equation of the parabola is
y = (x + 3) - 9.
Page #351 #1, 2b, 3d, 7cd.