Lesson 17: Quadratic Functions

𝐏𝐓𝐒 𝟑
Bridge to Calculus Workshop
Summer 2020
Lesson 17
Quadratic Functions
“A mathematician is a device for
turning coffee into theorems.”
- Paul Erdős -

Lehman College, Department of Mathematics
Quadratic Functions (1 of 4)
A quadratic function is any function of the form:
where 𝑎, 𝑏, and 𝑐 are real numbers, with 𝑎 ≠ 0. A
quadratic function is a polynomial function of degree 2.
The graph of a quadratic function is called a parabola.
Example 1. Graph the function 𝑓 𝑥 = 𝑥2
:
Solution. We first create a table with some values of 𝑥:
Next, we graph these points on the coordinate axes.
𝑥 −2 −1 0 1 2
𝑦
𝑓 𝑥 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐
0 11 44

Solution (cont’d). We create a graph from the table:
4
𝑥 𝑓(𝑥)
1
1
4
00
−1
−2
1
2
𝑓 𝑥 = 𝑥2

Parabolas are symmetric with respect to a line called
the axis of symmetry. The point where the axis
intersects the parabola is the vertex. When the leading
coefficient is positive, the graph of the parabola opens
upwards. Downwards for negative leading coefficient.

The simplest type of quadratic function is:
Its graph is a vertical parabola with vertex (0, 0).
When 𝑎 > 0, the vertex is the point with the minimum 𝑦-
value on the graph, and when 𝑎 < 0, the vertex is the
point with the maximum 𝑦-value on the graph:

Standard and Vertex Form (1 of 7)
The standard form of a quadratic function is given by:
The vertex form of a quadratic function is given by:
here 𝑎 ≠ 0, and ℎ, 𝑘 are the coordinates of the vertex.
Example 2. Find the vertex and axis of the parabola:
Solution. We complete the square in the formula:
+ 𝑏𝑥 + 𝑐
𝑓 𝑥 = 𝑎 𝑥 − ℎ 2
+ 𝑘
𝑓 𝑥 = 2𝑥2
+ 8𝑥 + 7
𝑓 𝑥 = 2𝑥2 + 8𝑥 + 7
= 2 𝑥2
+ 4𝑥 + 7
= 2 𝑥2
+ 4𝑥 + 4 − 4 + 7

Solution (cont’d). From the previous slide:
This is the vertex form. The vertex is:
The equation of the axis of symmetry is:
The orientation is:
Example 3. Find the vertex and axis of the parabola:
Solution. We complete the square in the formula.
𝑓 𝑥 = 2 𝑥2
+ 4𝑥 + 4 − 4 + 7
= 2 𝑥2
+ 4𝑥 + 4 + 2(−4) + 7
= 2 𝑥 + 2 2
− 8 + 7
= 2 𝑥 + 2 2
− 1
(−2, −1)
𝑥 = −2
𝑓 𝑥 = −𝑥2
+ 6𝑥 − 8
opens upwards

This is the vertex form. The vertex is:
The equation of the axis of symmetry is:
The orientation is:
= − 𝑥2 − 6𝑥 + 9 − 9 − 8
= − 𝑥2
− 6𝑥 + 9 − −9 − 8
= − 𝑥 − 3 2
+ 9 − 8
= − 𝑥 − 3 2 + 1
(3, 1)
𝑥 = 3
𝑓 𝑥 = −𝑥2
+ 6𝑥 − 8
= − 𝑥2
− 6𝑥 − 8
opens downwards

Example 4. Write the vertex form of the equation of the
parabola whose vertex is (1, 2) and that passes through
the point (3, −6):
Solution. The vertex form is given by the formula:
where ℎ, 𝑘 = (1, 2) and 𝑎 ≠ 0. It follows that:
Vertex form of equation:
𝑓 𝑥 = 𝑎 𝑥 − ℎ 2
+ 𝑘
𝑓 𝑥 = 𝑎 𝑥 − 1 2
+ 2 Write in vertex form
Substitute 𝑥 = 3 and 𝑓(𝑥) = −6−6 = 𝑎 3 − 1 2
+ 2
−6 = 4𝑎 + 2
−8 = 4𝑎
Simplify
Therefore: 𝑎 = −2
𝑓 𝑥 = −2 𝑥 − 1 2
+ 2

Example 5. Determine the vertex form of the following
quadratic function, where 𝑎 ≠ 0:
Solution. We factor the 𝑎, then complete the square.
𝑓(𝑥) = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐
= 𝑎 𝑥2
+
𝑏
𝑎
𝑥 +
𝑏
2𝑎
2
−
𝑏
2𝑎
2
+ 𝑐
= 𝑎 𝑥 +
𝑏
2𝑎
2
− 𝑎
𝑏
2𝑎
2
+ 𝑐
= 𝑎 𝑥2
+
𝑏
𝑎
𝑥 + 𝑐
𝑓(𝑥) = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐

It follows that:
𝑓 𝑥 = 𝑎 𝑥 +
𝑏
2𝑎
2
− 𝑎
𝑏
2𝑎
2
+ 𝑐
= 𝑎 𝑥 +
𝑏
2𝑎
2
− 𝑎
𝑏2
4𝑎2
+ 𝑐
= 𝑎 𝑥 +
𝑏
2𝑎
2
−
𝑏2
− 4𝑎𝑐
4𝑎
= 𝑎 𝑥 − ℎ 2 + 𝑘
ℎ = −
𝑏
2𝑎
𝑘 = −
𝑏2
− 4𝑎𝑐
4𝑎
and

Example 6. Use the formula to determine the vertex of
the following parabola:
Solution. The 𝑥-coordinate of the vertex is given by:
To find the 𝑦-coordinate, simply substitute 𝑥 = 2.
Therefore, the vertex is:
𝑓 𝑥 = 2𝑥2
− 8𝑥 + 3
ℎ = −
𝑏
2𝑎
= −
−8
2(2)
= 2
2 2 2
− 8 2 + 3𝑘 = 𝑓 2 =
= 2 4 − 16 + 3
= 8 − 16 + 3 = −5
(2, −5)

Graphs of Parabolas (1 of 8)
Example 7. Graph the parabola:
Solution.
Step 1. Determine the orientation:
Step 2. Find the 𝑦-intercept (𝑥 = 0):
Step 3. Determine the vertex:
Step 4. Start making a table of values for the function.
𝑓(𝑥) = −𝑥2
+ 4𝑥
opens downwards
𝑦 = − 0 2 + 4 0 = 0
ℎ = −
𝑏
2𝑎
= −
4
2(−1)
= 2
− 2 2 + 4 2 =𝑘 = −4 + 8 = 4

Perelman and Conway (1 of 1)
John H. Conway (1937-2020)
- English Mathematician
Grigori Perelman (b. 1966)
- Russian Mathematician

Step 4. Start making a table of values of the function.
Step 5. Do a quick sketch from the information known.
𝑥
𝑦
0
0
2
4

Step 6. Determine the 𝑥-intercepts (𝑦 = 0).
It follows that:
Add this information to the table:
Step 7. Plot more points as necessary:
Step 8. Graph the function from the information in the
table. Make sure to label the vertex and indicate the
axis of symmetry.
𝑥
𝑦
0
0
2
4
−𝑥2
+ 4𝑥 = 0
−𝑥(𝑥 − 4) = 0
𝑥 = 0 and 𝑥 = 4
0
4 −1
−5
5
−5

𝑥
𝑦
0
0
2
4 0
4 −1
−5
5
−5

Example 8. Graph the parabola:
Solution.
Step 1. Determine the orientation:
Step 2. Find the 𝑦-intercept (𝑥 = 0):
Step 3. Determine the vertex:
Step 4. Start making a table of values for the function.
𝑓 𝑥 = −2𝑥2
− 4𝑥 + 6
opens downwards
𝑦 = −2 0 2
− 4 0 + 6 = 6
ℎ = −
𝑏
2𝑎
= −
−4
2(−2)
= −1
−2 −1 2 − 4 −1 + 6 =𝑘 = −2 + 4 + 6 = 8

Step 4. Start making a table of values of the function.
Step 5. Do a quick sketch from the information known.
𝑥
𝑦
0
6
−1
8

Step 6. Determine the 𝑥-intercepts (𝑦 = 0).
It follows that:
Add this information to the table:
Step 7. Plot more points as necessary:
Step 8. Graph the function from the information in the
table. Make sure to label the vertex and indicate the
axis of symmetry.
𝑥
𝑦
0
6
−1
8
−2𝑥2
− 4𝑥 + 6 = 0
−2(𝑥2 + 2𝑥 − 3) = 0
𝑥 = 1 and 𝑥 = −3
0
1 −3
0
−2
6
−2(𝑥 − 1)(𝑥 + 3) = 0

Equations from Graphs (1 of 4)
Example 9. Write an equation in vertex form for the
parabola given below:

Solution. Find the coordinates of the vertex:
Find the coordinates of another point on the parabola,
preferably an 𝑥- or 𝑦-intercept:
The vertex form is given by the formula:
where ℎ, 𝑘 = (−2, −1) and 𝑎 ≠ 0. It follows that:
𝑓 𝑥 = 𝑎 𝑥 − ℎ 2
+ 𝑘
𝑓 𝑥 = 𝑎 𝑥 + 2 2
− 1 Write in vertex form
Substitute 𝑥 = 0 and 𝑓(𝑥) = 33 = 𝑎 0 + 2 2 − 1
3 = 4𝑎 − 1
4 = 4𝑎
Simplify
Therefore: 𝑎 = 1
𝑓 𝑥 = 𝑥 + 2 2
− 1
(−2, −1)
(0, 3)

Example 10. Write an equation in vertex form for the
parabola given below:

Solution. Find the coordinates of the vertex:
Find the coordinates of another point on the parabola,
preferably an 𝑥- or 𝑦-intercept:
The vertex form is given by the formula:
where ℎ, 𝑘 = (−2, 2) and 𝑎 ≠ 0. It follows that:
𝑓 𝑥 = 𝑎 𝑥 − ℎ 2
+ 𝑘
𝑓 𝑥 = 𝑎 𝑥 + 2 2
+ 2 Write in vertex form
Substitute 𝑥 = −1 and 𝑓(𝑥) = 00 = 𝑎 −1 + 2 2 + 2
0 = 𝑎 + 2
−2 = 𝑎
Simplify
Therefore: 𝑎 = −2
𝑓 𝑥 = −2 𝑥 + 2 2
+ 2
(−2, 2)
(−1, 0)

Range of a Quadratic Function (1 of 1)
Let 𝑓 𝑥 = 𝑎 𝑥 − ℎ 2
+ 𝑘 be a quadratic function in
vertex form. Then the domain of the function is all ℝ and
the range is given by:
Example 11. Find the range of 𝑓 𝑥 = −2𝑥2
+ 4𝑥 − 5.
Solution. Complete the square in the function formula:
Range:
[𝑘, ∞) if 𝑎 is positive
(−∞, 𝑘] if 𝑎 is negative
𝑓 𝑥 = −2𝑥2 + 4𝑥 − 5
= −2 𝑥2
− 2𝑥 − 5
= −2 𝑥2 − 2𝑥 + 1 − 1 − 5
= −2 𝑥 − 1 2
− 3
(−∞, −3] since: 𝑎 = −2 < 0 and 𝑘 = −3

Lesson 17: Quadratic Functions

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