U4L5: Finding the Vertex in Standard Form
Three Useful Forms for Quadratic Functions
Earlier this year when we studied Linear Functions, we found it convenient to have different forms of the
linear equations that fit the different data we were trying to model, As such, we used both slope-intercept
(y = mx + b) and point-slope (y - y1 = m(x - x1) ) forms. Similarly, there are different ways to write
quadratic equations. We have already seen standard and vertex forms, now we see the factored form too.
Standard Form
f ( x) = ax 2 + bx + c
Vertex Form
f ( x ) = a ( x − h) 2 + k
Factored Form
f (x) = a(x − m)(x − n)
this is the form you need for the
quadratic formula.
the point (h,k) is the vertex; a is a
vertical scaling factor.
m and n are the real zeros
(x-intercepts); a is a vertical
scaling factor
Finding the Vertex using Symmetry
Problem 1: A quadratic function is given as: 𝑓 𝑥 = 𝑥 ! + 2𝑥 − 3
a) Factor the function and find the x-intercepts.
b) Plot the x-intercepts you found above. Where
will the vertex be in relation to these points?
c) Find the x-value of the vertex.
d) Find the y-value of the vertex.
Problem 2: Repeat this process to find the vertex of the quadratic functions given below
a) 𝑓 𝑥 = 𝑥 ! + 3𝑥 − 28
b) 𝑓 𝑥 = 𝑥 ! − 6𝑥 + 8
U4L5: Finding the Vertex in Standard Form
Finding the Vertex by Using a Special Formula
We will use the next series of scaffolded problems to derive (figure out where something comes from) a
very useful formula for finding the vertex when you’re given the standard form: 𝑓 𝑥 = 𝑎𝑥 ! + 𝑏𝑥 + 𝑐
Problem 3
Step 1: By first expanding and then simplifying the general vertex form for a quadratic equation:
𝑓 𝑥 = 𝑎(𝑥 − ℎ)! + 𝑘, to show that it is equivalent to 𝑓 𝑥 = 𝑎𝑥 ! − 2𝑎ℎ𝑥 + 𝑎ℎ! + 𝑘
Step 2: By comparing your answer from above, 𝑓 𝑥 = 𝑎𝑥 ! − 2𝑎ℎ𝑥 + 𝑎ℎ! + 𝑘, and the standard form
of a quadratic function, 𝑓 𝑥 = 𝑎𝑥 ! + 𝑏𝑥 + 𝑐, state what a, b and c are equal to in terms of h and k.
Step 3: Hopefully you found that b is the coefficient of the x-value in the standard form of a quadratic
function. So, because 𝑓 𝑥 = 𝒂𝑥 ! + 𝒃𝑥 + 𝒄 = 𝒂𝑥 ! + (−𝟐𝒂𝒉)𝑥 + (𝒂𝒉𝟐 + 𝒌), b = -2ah. Solve this
equation for h.
Solution: Because h is known to be the x-coordinate of the vertex, we could also say the x-coordinate of
!
the vertex is: − !! when a quadratic function is given in standard form: 𝑓 𝑥 = 𝑎𝑥 ! + 𝑏𝑥 + 𝑐.
Use this new formula to find the vertex of the function: 𝑓 𝑥 = 𝑥 ! + 2𝑥 − 3
The formula for the vertex of a parabola when given an equation in standard form is:
x – coordinate of the vertex: _______________________________
Equation for the line of symmetry: ______________________________
U4L5: Finding the Vertex in Standard Form
Group Exploration Exercises
1. Find the following for each of the quadratic equations given below.
a)
b)
c)
d)
e)
Find the y-intercept
Find the x-intercepts by factoring.
Find the x-coordinate of the vertex.
Find the y-coordinate of the vertex.
Sketch an accurate graph of the function including at
least 5 points.
a. f (x) = x 2 − 2x − 3
2
b. f (x) = x + 7x +10
2
c. f (x) = x − 6x
U4L5: Finding the Vertex in Standard Form
2. Read the example below, then answer the questions that follow.
Example: Find the equation of a quadratic function with zeros (1,0)
and (4,0) that passes through the point (0,–5).
y
3
2
1
Wha?!?!? As you can see, there are lots of quadratics that could have
the zeroes of (1,0) and (4,0), but exactly one of those can also pass
through the point (0,–5).
−3
x
−2
−1
1
2
3
4
5
−1
−2
Here is how we find the equation:
 Since we have the zeros, use the “factored form” of the quadratic
y = a(x − m)(x − n)
(general equation)
y = a( x − 1)( x − 4)
(plug in the known zeroes)
−3
−4
−5
‚ To specify a unique quadratic, find the scaling factor “a” (has it been vertically stretched or flipped
across the x-axis?) using the extra point.
y = a( x − 1)( x − 4) , with the point (0,–5)
− 5 = a(0 − 1)(0 − 4)
(substitute for x and y)
− 5 = a(−1)(−4)
(simplify)
− 5/ 4 = a
(solve for a)
ƒ Put it all together: y = (−5 / 4)( x − 1)( x − 4) .
In each of the following problems, you are given some information about a quadratic function. Write the
equation of each quadratic, expressing your final answer in either standard form or vertex form. Hint:
Use the form that is the most helpful.
a. vertex (5,2) and through the point (−2,1).
b. zeros (3,0) and (−5,0) and through point (−1,3)
6
U4L5: Finding the Vertex in Standard Form
c. Only x-intercept is (5,0) and y-intercept of 3.
d. Has the graph given on the right.
e. Has values as follows: f(2) = 0, f(0) = 3, f(−3) = 0.