# a + b

```Name: _________________________________________________________ Date: ______________
REVIEW: Graphing Linear and Quadratic Functions
Pre-Calculus
Linear Function
A linear function is one that can be written in Slope-Intercept form
f(x) = mx + b
Function form
Example: f(x) = 3x - 1 m = 3, b = -1
y = mx + b
Equation form
Example: y = 3x - 1 m = 3, b = -1
where m and b are fixed numbers (the names m and b are traditional).
Think about linear functions as transformations of the Identity Function, y = x.
What are the transformation being performed for each of the equations?
1.
2.
3.
4.
y = 3x – 1
y = 2x + 2
y=&frac12;x–5
y=-x+3
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Slope of a Line
The slope of a line is given by the ratio
m
y change in y rise


 slope
x change in x run
Linear Equations
Computing the slope of a line through 2 point (x1,y1) and (x2,y2) we use the slope
formula:
y y 2  y1
m

x x2  x1
Example:
Find the slope of the line through (1, 3) and (5, 11)
y y 2  y1 11  3 8
m


 2
x x2  x1 5  1 4
5. Find the slope of the line through (-1, 0) and (2, -2) ___________________
Finding the Equation of a Line
Problem: Find the equation of the line that passes through the point (-2, 5) and has a slope of -4.
Solution:
 Substitute y1 , x1 and m in the point slope form of a line
y - y1 = m(x - x1)
y - 5 = - 4(x - (-2))
y = - 4x - 3
6. Find the equation of the line that passes through the point (3 , 0) and has a slope of – 1.
Problem: Find the equation of the line that passes through the points (0 , -1) and (3 , 5).
Solution:
 We first calculate the slope of the line
m = (5 - (-1)) / (3 - 0) = 6 / 3 = 2
 Use the slope and any of the two points to write the equation of the line using the point
slope form.
y - y1 = m(x - x1)
using the first point
y - (-1) = 2(x - 0)
y = 2x - 1
7. Find the equation of the line that passes through the points (2 , 0) and (3 , 3).
A quadratic function of the variable x is a function that can be written in the form
f(x) = ax2 + bx + c
Function form
y = ax2 + bx + c
Standard form (an equation)
where a, b, and c are fixed numbers (with a ≠ 0).
Examples
1. f(x) = 3x2 - 2x + 1
2. g(x) = -x2
3. h(x) = 3x + 1
a = 3, b = -2, c = 1
a = -1, b = 0, c = 0
Not a quadratic function because a = 0 (it’s a linear function)
GCF
Greatest Common Factor
Binomials
Trinomials
2 Perfect
Squares
Difference of 2 Squares
Factor it
Un-FOIL-ing
a2 – b2 = (a + b)(a – b)
Factor it
Box Method
Factor it
x
 b  b 2  4 ac
2a
Features of a Parabola
A parabola in standard form f(x) = ax2 + bx + c (a ≠ 0) has the following features:
1) Concavity – does it open up or down
2) The axis of symmetry (the line of symmetry for the parabola)
3) The vertex (the minimum or maximum point of the parabola)
4) The y-intercept (where it crosses the y-axis)
5) The x-intercepts (where it crosses the x-axis)
To graph a quadratic function, you must find each of the features listed above.
EXAMPLE: Graph the function f(x) = x2 - 5x + 6
1. Concavity
If a &gt; 0, the parabola is concave up [opens up];
if a &lt; 0, the parabola is concave down [opens down].
Example
For f(x) = x2 - 5x + 6 has a = 1 &lt; 0, so the graph is concave up (opens up)
2. Axis of Symmetry
A vertical line with the equation x 
x=
-b -(-5) 5
=
=  2.5
2a 2(1) 2
b
2a
3. Vertex
b
(same as the axis of
2a
b
symmetry, which was found in #2 above). The y-coordinate is f(
).
2a
A point, (x, y). The x-coordinate of the vertex is
Example
The graph of f(x) = -3x2 - 6x - 3 has vertex with the following coordinates:
x-coordinate=
-b
=2.5
2a
 b 
2
y  coordinate  f(x)  f 
  f(2.5)  (2.5)  5(2.5)  6  .25
2
a


The vertex is (-1, -3)
y-Intercept - where the graph crosses the y-axis. Find it by setting x = 0.
The y-intercept is given by y = c.
Example
The graph of f(x) = -3x2 - 6x - 3 has c = -3, so the y-intercept is given by y = -3.
x-Intercepts - where the graph crosses the x-axis. Find it by setting y = 0.
The x-intercepts, if they exist, are given by setting f(x) or y equal to zero:
ax2 + bx + c = 0
and solving for x. To do this, you will need to factor the (see flow chart above).
Note: If the quadratic has no real factors (i.e. the square-root yields imaginary numbers)
then there are no x-intercepts -- the parabola is entirely above or below the x-axis.
Example
Find the x-intercepts of f(x) = x2 - 5x + 6 by solving the quadratic
x2 - 5x + 6 = 0
step 1: set f(x) = 0
To solve for x, factor the quadratic:
x2 - 5x + 6 = 0
(x - 3)(x - 2) = 0
x–3=0
x–2=0
x=3
x=2
The x-intercepts are 3 and 2. These are where the graph crosses the x-axis. These
are the points (3, 0) and (2, 0).
To Graph a Quadratic Function: Draw a faint dotted line for the axis of symmetry (it is a
vertical line). Place dots on the grid for the vertex (should lie on the axis of symmetry),
and the x- and y-intercepts. If possible, use symmetry to find matching points. Find
other points with a T-table if necessary (you should have 5 points at a minimum).
Connect the dots with a smooth graph. Put arrow heads on the ends.
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