Download U4L4: Standard Form of Quadratic Functions Turning Vertex Form

U4L4: Standard Form of Quadratic Functions
Turning Vertex Form into Standard Form
Last class we played with the graphing or vertex form of a quadratic function: 𝑓 𝑥 = 𝑎 𝑥 − ℎ ! + 𝑘.
Recall that we viewed all quadratics in vertex form as a transformation of the parent function f(x) = x2. Also
recall, that we call it vertex form, because in this form we can easily and immediately know the vertex of
the graph. What is the location, in general, of the vertex of a parabola written as: 𝑓 𝑥 = 𝑎 𝑥 − ℎ ! + 𝑘?
However, we don’t always see equations for quadratics in this form. Today we’ll look at another common
form to write quadratics, standard form: 𝑓 𝑥 = 𝑎𝑥 ! + 𝑏𝑥 + 𝑐 (look familiar?!?)
Active Example 1: Expand and simplify the following quadratic equation in vertex form to show that
f (x) = (x +1) 2 − 4 is equivalent to the expression f (x) = x 2 + 2x − 3 in standard form.
!
Active Example 2: Now do the same for 𝑔 𝑥 = ! (𝑥 + 2)! − 8. Expand and simplify to show that it is
!
equivalent to 𝑔 𝑥 = ! 𝑥 ! + 2𝑥 − 6.
You have now discovered and new form of quadratic equations, this is called:
__________________ form, written as y = ____________________________
Now, let’s explore what this form of quadratic equation can tell us…
U4L4: Standard Form of Quadratic Functions
We hope that you recognize this form of quadratic equations from our work with diamond puzzles. These
diamonds help us to factor the quadratic expressions, which become extremely useful when finding the
________________________ of the parabola.
Example 3: Use a diamond puzzle to factor and solve f(x) = 0 for the
function below. Then plot the solutions on the graph.
f (x) = x 2 + 2x − 3
What points have you found?
Where will the vertex be in relation to your solutions to the problem above?
What is the x-value of the vertex?
Use this x-value to find the y-value of the vertex.
Graph the function to verify your answers to the questions posed above.
!
Example 4: First factor ! out of the polynomial, then use a diamond
puzzle to factor and solve f(x) = 0 for the function below. Finally, plot the
solutions on the graph.
!
𝑔 𝑥 = ! 𝑥 ! + 2𝑥 − 6
What points have you found?
Where will the vertex be in relation to your solutions to the problem above?
What is the x-value of the vertex?
Use this x-value to find the y-value of the vertex.
Graph the function to verify your answers to the questions posed above.
U4L4: Standard Form of Quadratic Functions
1. Groups Explore. Notice from the previous page that the x-value for the axis of symmetry is halfway
between the x-intercepts. The x-value of the axis of symmetry is, in fact, midway between any two
points with the same y-value, due to the symmetrical nature of parabolas.
The table below gives some information about some other parabola graphs. Fill in the blanks with the
missing information. A sketch is often very helpful.
x-intercepts
axis of symmetry
Vertex
a.
x = 1 and x = 7
x = _____
(_____, 6)
b.
x = –5 and x = 3
x = _____
(_____, 2)
c.
x = 1 and x = _____
x=5
(_____, –3)
d.
x = _____ and x = 8
x = _____
(3, 4)
e.
x = _____
x=2
(2, 0)
f.
x = _____ and x = 6
x=0
(_____, –4)
sketch
U4L4: Standard Form of Quadratic Functions
2. For the following quadratics in standard form, factor to find the zeros, then graph the parabola, showing
at least 5 points.
a. 𝑓 𝑥 = 𝑥 ! − 2𝑥 − 3
b. 𝑓 𝑥 = 𝑥 ! + 7𝑥 + 10
c. 𝑓 𝑥 = 𝑥 ! − 6𝑥
d. 𝑓 𝑥 = 𝑥 ! − 4
3. For the following graphs, use the x-intercepts to write the equation in
factored form, 𝑦 = 𝑎(𝑥 − 𝑚)(𝑥 − 𝑛), then expand to write the equation in standard form.