WHAT YOU KNOW: We used the rectangular coordinate system to represent ordered pairs of real numbers and to graph equations in two variables. We saw that linear equations can be written in the form a x + b = 0 , a ≠ 0 , and quadratic equations can be written in the general form a x 2 + b x + c = 0 , a ≠ 0 . We solved linear equations. We saw that some equations have no solution, whereas others have all real numbers as solutions. We solved quadratic equations using factoring, the square root property, completing the square, and the quadratic formula. We saw that the discriminant of a x 2 + b x + c = 0 , b 2 − 4 a c , determines the number and type of solutions. We performed operations with complex numbers and used the imaginary unit i ( i = − 1 , where i 2 = − 1 ) to represent solutions of quadratic equations with negative discriminants. Only real solutions correspond to x -intercepts. We also solved rational equations by multiplying both sides by the least common denominator and clearing fractions. We developed a strategy for solving a variety of applied problems, using equations to model verbal conditions. In Exercises 1-12, solve each equation. ( x + 3 ) 2 = 24
WHAT YOU KNOW: We used the rectangular coordinate system to represent ordered pairs of real numbers and to graph equations in two variables. We saw that linear equations can be written in the form a x + b = 0 , a ≠ 0 , and quadratic equations can be written in the general form a x 2 + b x + c = 0 , a ≠ 0 . We solved linear equations. We saw that some equations have no solution, whereas others have all real numbers as solutions. We solved quadratic equations using factoring, the square root property, completing the square, and the quadratic formula. We saw that the discriminant of a x 2 + b x + c = 0 , b 2 − 4 a c , determines the number and type of solutions. We performed operations with complex numbers and used the imaginary unit i ( i = − 1 , where i 2 = − 1 ) to represent solutions of quadratic equations with negative discriminants. Only real solutions correspond to x -intercepts. We also solved rational equations by multiplying both sides by the least common denominator and clearing fractions. We developed a strategy for solving a variety of applied problems, using equations to model verbal conditions. In Exercises 1-12, solve each equation. ( x + 3 ) 2 = 24
Solution Summary: The author explains how to calculate the solution of the equation (x+3)2=24.
WHAT YOU KNOW: We used the rectangular coordinate system to represent ordered pairs of real numbers and to graph equations in two variables. We saw that linear equations can be written in the form
a
x
+
b
=
0
,
a
≠
0
, and quadratic equations can be written in the general form
a
x
2
+
b
x
+
c
=
0
,
a
≠
0
. We solved linear equations. We saw that some equations have no solution, whereas others have all real numbers as solutions. We solved quadratic equations using factoring, the square root property, completing the square, and the quadratic formula. We saw that the discriminant of
a
x
2
+
b
x
+
c
=
0
,
b
2
−
4
a
c
, determines the number and type of solutions. We performed operations with complex numbers and used the imaginary unit
i
(
i
=
−
1
,
where
i
2
=
−
1
)
to represent solutions of quadratic equations with negative discriminants. Only real solutions correspond to x-intercepts. We also solved rational equations by multiplying both sides by the least common denominator and clearing fractions. We developed a strategy for solving a variety of applied problems, using equations to model verbal conditions.
In Exercises 1-12, solve each equation.
(
x
+
3
)
2
=
24
Formula Formula A polynomial with degree 2 is called a quadratic polynomial. A quadratic equation can be simplified to the standard form: ax² + bx + c = 0 Where, a ≠ 0. A, b, c are coefficients. c is also called "constant". 'x' is the unknown quantity
Name:
Tay Jones
Level Two
Date:
Algebra 3 Unit 3: Functions and Equations Practice Assessment
Class:
#7-OneNote
1. The function f(x) = x² is transformed in the following functions. List the vertex for each function,
circle whether the function opens up or down, and why. All three parts must be correct to receive Level 2
points. You can receive points for a, b, and c.
a) g(x) = -2(x+5)²
Vertex:
Opens Up Opens Down
Why? ais negative
-2
Vertex:
b) g(x) = (x + 2)² - 3
c) g(x) = -4(x + 2)² + 2
Opens Up
Opens Down
Vertex:
Opens Up
Opens Down
Why?
4
Ca is negative)
Why? his positive
2. The graph of the function f(x) is shown below. Find the domain, range, and end behavior. Then list the
values of x for which the function values are increasing and decreasing.
f(x)
Domain:
End Behavior:
As x → ∞o, f(x) -> -6
As x, f(x) ->
Range:
Where is it Increasing? (002]
Where is it Decreasing? (1,00)
Show what to do on the graph visually please!
The county's new asphalt paving machine can surface 1 km of highway in 10 h. A much older machine can surface 1 km in 18 h. How long will it take them to surface 21 km of highway if they start at opposite ends and work day and night?
Chapter 1 Solutions
Algebra And Trigonometry 6th. Edition Annotated Instructor's Copy Blitzer
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.
Linear Equation | Solving Linear Equations | What is Linear Equation in one variable ?; Author: Najam Academy;https://www.youtube.com/watch?v=tHm3X_Ta_iE;License: Standard YouTube License, CC-BY