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AA2-9 graphing absolute value functions A-level

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To further explore these transformations and how to write an exact equation given a graph. Do the Desmos activity AA2-9 Graphing absolute values A-level. The link is in the Canvas module.

In this lesson we are going to investigate absolute value equations in the graphing form of f(x) = a|x-h| + k, we will determine how numbers at a, h and k transform the parabola.

The first thing we need to know is the shape of an absolute value function.

Pitanje 1
1.

Graph the parent function f(x) = |x| by using a table and evaluating f(-3), f(-2), f(-1), f(0), f(1), f(2), f(3).

Objavili smo novi i poboljšani tip pitanja za grafički prikaz! Studenti više neće moći odgovoriti na ovo pitanje.
Pitanje 2
2.

What is the slope of the right half of the graph where x > 0?

Pitanje 3
3.

What is the slope of the left half of the graph where x < 0?

Pitanje 4
4.

We will call the point where the two halves meet (the point of the V shape), the vertex.

Given the equation f(x) = |x + 7| - 4. Where is the vertex of f(x)?

Pitanje 5
5.

Check you answer above by graphing f(x) = |x + 7| - 4. The absolute value symbol of a vertical line "|" is on most keyboards, or you can use the keyboard button.

Objavili smo novi i poboljšani tip pitanja za grafički prikaz! Studenti više neće moći odgovoriti na ovo pitanje.
Pitanje 6
6.

Now we need to determine how "a" transforms the graph of a absolute value equation.

Graph the function f(x) = 2|x|

Objavili smo novi i poboljšani tip pitanja za grafički prikaz! Studenti više neće moći odgovoriti na ovo pitanje.
Pitanje 7
7.

Describe how an "a" value of 2 transforms the parent graph.

Pitanje 8
8.

Graph f(x) = -3|x|

Objavili smo novi i poboljšani tip pitanja za grafički prikaz! Studenti više neće moći odgovoriti na ovo pitanje.
Pitanje 9
9.

Describe how an "a" value of -3 transforms the parent graph.

Pitanje 10
10.

Match the graph and its equation.

Stavka koja se može prevućiarrow_right_altOdgovarajuća stavka

f(x) = -|x - 3| -1

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f(x) = 4|x - 3| - 1

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f(x)=3|x - 3| - 1

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f(x) = -2|x - 3| -1

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