Algebra 1 Chapter 4 power point

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Transcript Algebra 1 Chapter 4 power point

Chapter 4
Two points
determine a
line
 Standard Form
 Ax + By = C
 Find the x and
the y-intercepts


An equation represents an
infinite number of points
in a relationships
 When given an equation,
 make a T-chart
 substitute the domain (x
values) and find the
corresponding range (yvalues)
 A point and a slope can name
a line
 y = mx + b
 plot the y-intercept
 use the slope to find more
points
 Y = 3/4 x – 2
 3x + 2y = 6
 -2x + 5y = 10
 2y = 1

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Increasing
Decreasing
Zero Slope

Undefined
Slope or NO
slope
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y = (positive number)x + b
y = (negative number)x + b
y = constant (domain is all
real numbers and the
range is the constant)
X = constant (a vertical line
is not a function so there is
no y-intercept form for it

Change the intercept
 Y=x+7
 y=x+5
 Y=x–1
 Y=x–¾
Change the slope

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Y = 1/3x
y = 4x
Y = 10x
Y = -5x

Change both
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Y = 1/3x + 7
Y = -3/4x -5
Y = 8x -2
Y = -4x – 3
Y = 5/6x + 9
WHEN GIVEN A POINT
AND A SLOPE (NOT THE
Y-INTERCEPT)

Given:
 Pt (2,1) and slope 3
 Pt((4, -7) and slope 1
 Pt ((2,-3) and slope
1/2
WHEN GIVEN TWO
POINTS

Given:
 (3,1), (2,4)
 (-1, 12), (4, -8)
 (5,-8), (-7, 0)
 Given point (3,-2)
and slope ¼
 Given point (-2, 1)
and slope -6
 y + 2 = ¼(x – 3)
 y –(-2)= ¼(x – 3)
 y – 1 = -6(x + 2)
 y – 1 = -6(x –(-2))
 y – y1 = m(x – x1), where (x1, y1) is a
specific point
 Where does this equation come
from?
 m = y1 – y2
x1 – x2
Standard Form
Slope- Intercept
 Point-Slope


Ax +By = C
y = mx + b
 y – y1= m(x-x1)
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Find the equation in:
 Point slope form
 Standard form
 Slope-intercept form

You need to know how to identify key
elements from each type of equation and
when to use each!
 y = 2x – 4
 y = -3/4x + 3
y=½x–7
 y = -1/2 x + 2
 y = -2x + 5
 y = -3/4 x
 y = -3x + 4
 y = 4/3 x – 1
 y = 2x + 5
 y = .5x - 3
Parallel lines have the same slope
Write an equation for a line that passes
through the point (-3, 5) parallel to the
line y = 2x - 4
Write and equation for a line passing
through the point (4,-1) and parallel to the
line y = ¼ x + 7


 Intersecting lines have different
slopes
 Write an equation for a line that
intersects the line y = -2/3 x + 5 and
goes point (-1, 3)
 Write an equation for a line that
intersects the line 3x – 4y = 10
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The slopes of perpendicular lines are opposite
reciprocals
Write and equation for a line that passes
through the point (-4,6) and is perpendicular
to the line 2x + 3y = 12
Write an equation to a line that passes
through the point (4,7) and is perpendicular
to the line y = 2/3 x - 1
Bivariate Data
Regression Lines (line of best fit)
Correlation
 Causation
 Correlation coefficient (r factor)
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 Additive Inverse (opposite)
 Multiplicative Inverse (reciprocal)
 Square Root (undoes squaring)
 Solving Equations

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If one relation contains the element (a,b), then
the inverse relation will contain the element
(b,a)
EX:
A
B
(-3, -6)
(-6, -3)
(-1, 4)
(4, -1)
(2, 9)
(9, 2)
((5, -2)
(-2, 5)
~Display as a set of ordered pairs, Table, Mapping,
Graph
 “Mathalicious example”~ wins per
million we reversed to millions per
win
 y= x + 3
 y =2x + 3
 y = -1/3x + 2
 y = -3/4x -1

To find the inverse function f-1 (x) of the linear
function f(x), complete the following steps:
 Step 1~ Replace f(x) with y in the
equation f(x)
 Step 2~ Interchange y and x in the
equation
 Step 3~ Solve the equation for y
 Step 4~ Replace y with f-1 (x) in the
new equation
 f(x) = 4x – 6
 f-1(x) = x + 6
 f(x) = -1/2x + 11
4
 f-1(x) = -2x +22
 f(x) = -3x + 9
 f(x) = 5/4x – 3
 f-1(x) =
-1/3x +3
 f-1(x) = 4/5 x + 12/5



Mathalicious example”~ wins per million we
reversed to millions to win
 f(x)= .103x – 2.96 (NFL cost verses wins)
 F-1(x) = 9.7x + 2.87 (NFL wins verses cost)
Celsius verse Fahrenheit
 C(x) = 5/9(x – 32
 C-1(x) = F(x) (Fahrenheit)
Car rental cost per day
 C(x) = 19.99 + .3x
 C-1 (x) = total number of miles