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Playlist Algebra and Functions Test Questions 74 videos

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SAT Math 3.4 Algebra and Functions
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SAT Math 3.4 Algebra and Functions

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SAT Math 9.2 Algebra and Functions
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SAT Math 9.2 Algebra and Functions

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SAT Math 9.3 Algebra and Functions
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SAT Math 9.3 Algebra and Functions

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SAT Math 9.3 Algebra and Functions 225 Views


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Description:

SAT Math 9.3 Algebra and Functions

Language:
English Language

Transcript

00:05

The line y = -2x + 4 is reflected over the y-axis, then translated up three units and left two units.

00:12

What is its new y-intercept?

00:17

There are two ways to do this problem: algebraically and graphically.

00:24

Let’s start with “algebraically.”

00:26

We’re given a line and a series of transformations, so we’ll apply them to the equation of the line.

00:32

First is a reflection over the y-axis. That means we have to change x into negative x.

00:38

Our line changes to y = 2x + 4. Next up is a “3 unit up” translation.

00:45

For this, we just… add 3 to the line. We get y = 2x + 7.

00:50

Finally, we have a left shift of 2.

00:53

To apply this one, add 2 to x.

00:56

To get this equation back into slope intercept form, we distribute the 2 into the parentheses

01:00

and get y = 2x + 11.

01:03

Remember that the y intercept of a line in slope intercept form is just the constant

01:07

added onto the end.

01:09

In this case… 11. That's our y intercept.

01:13

Another way to solve this problem is graphically.

01:15

We start with the graph of y = -2x + 4

01:20

First, we have the reflection across the y-axis.

01:23

Then we move it up 3.

01:25

Finally, we shift the graph 2 to the left.

01:28

By looking at the graph, we can also see that the y intercept is 11.

01:33

Booyah. Two paths – same destination.

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