# Basic Algebra - Relations and functions

#28: y=-2x2+3

x y=-2x2+3 y

0 y=-2(0)2+3 3

1 y=-2(1)2+3 1

2 y=-2(2)2+3 -5

-1 y=-2(-1)2+3 1

-2 y=-2(-2)2+3 -5

• The graph of the relation is (x)=-2x2+3.

• My 5 points for this equation is as follows:

(0,3)

(1,1)

(2,-5)

(-1,1)

(-2,-5)

• The key points and general shape of this graph is upward and the vertex is 3 located on the y-intercept. This is a parabola.

• The domain is (-,).

• The range is (-, 3].

• This graph is a function because each y value equals one x value. "If the value of the variable y is determined by the value of the variable x, then y is a function of x. So is a function of means is uniquely determined by" (Dugopolski, 2012, p. 690).

• The vertical line test passes through any x value and it only crosses that line once.

#36: y=2(x)+1

x y=2(x)+1 y

0 y=2(0)+1 1

1 y=2(1)+1 3

2 y=2(2)+1 3.82

4 y=2(4)+1 5

6 y=2(6)+1 5.89

• The graph of the relation is (x)=2(x)+1.

• My 5 points for this equation is as follows:

(0,1)

(1,3)

(2,3.82)

(4,5)

(6,5.89)

• This graph is a function because each y value equals one x value.

• The general shape of this graph is a curve that plateau. There is no vertex.

• I selected problem 36 to be shifted 3 units upward and 4 units to the left.

y=2(x)+1

y=2(x+4)+1+3

y=2(x+4)+4

• This is the transformation of the function. I added 3 outside the radical and added 4 inside the radical.

Page 709 #28

This problem is an example of a quadratic function. We know this because it is in the form f(x)=ax2+bx+c where a, b, and c are real numbers and a is not equal to 0. In this case, we will graph the given function by plotting enouh points to figure out the shape of the graph.

Y = -2x2+3 original equation.

X Y In this portion, I just plugged in the values under the x column

-2 -5 into the equation above and found the corresponding solutions

-1 1 for y. These are the ordered pairs that will appear on the

0 3 graph.

1 1

2 -5

This is a function because each value for y only has one x value and passes a vertical line test. The vertex for this parabola is [0,3], the domain is {[-∞, ∞]}, and the range is {[-∞,3]}.

Page 709 #36

This problem is a square root function because it comes in the form f(x)= √x.

Y = 2√(x)+1 original equation.

X Y In this portion, I just plugged in the values under the x column

0 1 into the equation above and found the corresponding solutions

1 3 for y. These are the ordered pairs that will appear on the graph.

4 5

9 7

This is a function because each value for y only has one x value and passes a vertical line test. The domain for this graph is {[0, ∞]} and the range is [{1, ∞}].

Had this line been shifted up three and left four a transformation in the graph and equation would occur. (x)+ c represents a line shifting up c units and (x+c) represents a line shifting left c units. The change to the equation is shown below.

Y= 2√(x )+1

Y= 2√(x+4) +1+3

Y= 2√(x+4) +4

#### Solution Preview

Relations and Functions

#28: y=-2x2+3

x y=-2x2+3 y

0 y=-2(0)2+3 3

1 y=-2(1)2+3 1

2 y=-2(2)2+3 -5

-1 y=-2(-1)2+3 1

-2 y=-2(-2)2+3 -5

• The graph of the relation is (x)=-2x2+3.

• My 5 points for this equation is as follows:

(0,3)

(1,1)

(2,-5)

(-1,1)

(-2,-5)

• The key points and general shape of this graph is an downward parabola and the vertex is at x=3 located on the y-intercept. The point of the vertex is (0,3)

• The domain is (-,).

• The range is (-,3].

• This graph is a function because each y value equals one x value. "If the value of the variable y is determined by the value of the variable x, then y is a function of x. So is a function of means is uniquely determined by" ...

#### Solution Summary

Relations and functions for basic algebra are provided.