Radical Functions & Graphing

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Radical Functions & Graphing

Graph 𝑦 = √π‘₯ using a table of values and state the domain and range.

Domain: Range:

Notice the domain is restricted. Why? What is an easy way to obtain the domain?

Radical functions can be graphed using transformations – just like quadratic functions! For example, 𝑦 = π‘Žβˆšπ‘₯ βˆ’ β„Ž + π‘˜. The vertex is (β„Ž, π‘˜), and the vertical expansion/compress is π‘Ž. When graphing parabolas, what happened to 𝑦 = π‘₯2 when it became 𝑦 = βˆ’π‘₯2 (ie. When the π‘Ž value was -1)?

This is called a vertical reflection (a reflection through the _________________________). Therefore, graph 𝑦 = βˆ’βˆšπ‘₯ without a table of values on the grid above, and then state the domain and range below:

What is the β€˜basic count’ for a radical graph? Right ____ Up _____ Remember, the π‘Ž value alters the up/down count.

Graph 𝑦 = 3√π‘₯ and 𝑦 = βˆ’12√π‘₯ on the graph below & state the domain, range, x-int, y-int x y

To find the x-intercept, set y = 0.

To find the y-intercept, set x = 0.

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Graph 𝑦 = √π‘₯ + 3 + 2 and 𝑦 = βˆ’2√π‘₯ βˆ’ 1 βˆ’ 4. Give D, R, x-int, y-int for each.

So, using transformation theory that was learned for quadratics, we can graph other types of functions with the same principles.

The graph 𝑦 = βˆ’βˆšπ‘₯ is a reflection through the horizontal line through the vertex of the 𝑦 = √π‘₯ graph. Using a table of values, graph 𝑦 = βˆšβˆ’π‘₯ below, and state the domain and range.

D: R:

On the grid above, graph 𝑦 = βˆ’βˆšβˆ’π‘₯ without a table of values, and state the domain and range.

For the following radical functions, state the domain, range, x-int, and y-int. a) 𝑦 = βˆ’4√π‘₯ βˆ’ 3 + 8 b) 𝑦 =12√π‘₯ + 5 βˆ’ 1

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Graph the following functions. A good way to do this is to locate the vertex first, and then establish the β€˜count’ from the vertex by altering the β€˜basic count’ accordingly.

a) 𝑦 = βˆ’4√π‘₯ βˆ’ 3 + 8 b) 𝑦 =12√4 βˆ’ π‘₯ βˆ’ 1

Here is a summary of transformations for radical graphs:

Graph the following and state the domain, range, x-intercept, and y-intercept 𝑦 βˆ’ 6 = βˆ’2√3 βˆ’ π‘₯

π’š = Β±π’‚βˆšΒ±(𝒙 βˆ’ 𝒉) + π’Œ Steps for graphing:

1) If x is negative, factor it out. 2) Locate the vertex (h, k)

3) From the vertex, build the count:

If no negative in front in radical, go right. If negative in front in radical, go left.

If a is positive, count up. If a is negative, count down. Multiply up/down count by a value. Basic Count: right 1, up 1

right 4, up 2 right 9, up 3

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Graphing π’š = 𝒇(𝒙) 𝒂𝒏𝒅 π’š = βˆšπ’‡(𝒙)

The graphs of 𝑦 = 𝑓(π‘₯) and 𝑦 = βˆšπ‘“(π‘₯) could have different domains & ranges because of the restriction on an even root function.

Graph 𝑦 = π‘₯2 and 𝑦 = √π‘₯2 and state the domain and range.

Graph 𝑦 =1

2(π‘₯ + 2)

2βˆ’ 8 and 𝑦 = √1

2(π‘₯ + 2)

2βˆ’ 8 and state the domain and range.

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Absolute Value Equations What is the absolute value of a number?

Example 1 – Simplify a) |6| b) |βˆ’4| c) |βˆ’73

|

Example 2 – Solve |π‘₯| = 2

Steps for solving an absolute value equation:

1) Get the absolute value by itself on one side (everything not in the absolute value should be on the other side).

2) Set up two cases: the positive case and the negative case. Solve for each case. 3) Check each solution to see if it is an actual or extraneous solution.

Example 3 – Solve |π‘₯ + 7| = 10 First, by inspection:

Example 4 – Solve |π‘₯ βˆ’ 2| + 3 = 9

Positive Case: Negative Case: Check:

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Example 5 – Solve |3π‘₯ βˆ’ 2| = 1 βˆ’ π‘₯ algebraically

Example 6 – Solve |π‘₯ βˆ’ 3| + 7 = 4

Example 7 – Solve |4π‘₯ βˆ’ 5| + 2 = 2

Check:

Check:

An Absolute Value Equation with No Solution:

no solutions

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Example 8 – Solve |π‘₯ + 5| = 4π‘₯ βˆ’ 1 algebraicially Example 9 – Solve |π‘₯2βˆ’ 7π‘₯ + 2| = 10 Example 10 – Solve |π‘₯ βˆ’ 2| = π‘₯ βˆ’ 2 Check: Check: quadratic absolute value equations

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Absolute Value Functions & Graphing Example 1 – Graph 𝑦 = |π‘₯|

At times, absolute value functions are set up analogous to quadratic functions in standard form, using h, k, and a values to graph the function.

What is the basic count?

Summarize how each component of the function affects the graph:

𝑦 = Β±π‘Ž|π‘₯ βˆ’ β„Ž| + π‘˜ & 𝑦 = Β±π‘Ž|βˆ’(π‘₯ βˆ’ β„Ž)| + π‘˜

Example 1 – Graph each function and state the D, R, x-int, y-int:

a) 𝑦 = |π‘₯| βˆ’ 3 b) 𝑦 = |π‘₯ + 4| x y -3 -2 -1 0 1 2 3 The graph 𝑦 = |π‘₯| consists of two graphs:

This is why an absolute value graph is called a piecewise function.

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c) 𝑦 = βˆ’2|π‘₯| d) 𝑦 = βˆ’1

2|1 βˆ’ π‘₯| + 2

e) 𝑦 + 7 = 3|2 βˆ’ π‘₯| f) 𝑦 = 3 βˆ’14|π‘₯ βˆ’ 2|

Look at (a) above. The right branch only would have the equation _______________ and the left branch only would have the equation _____________. Thus we can write

𝑦 = |π‘₯| βˆ’ 3 as a piecewise function:

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If a graph is constructed for any function 𝑓(π‘₯), what will the graph for |𝑓(π‘₯)| look like? Example 3 – Graph 𝑦 = π‘₯2βˆ’ 4 using a table of values. Then graph 𝑦 = |π‘₯2βˆ’ 4|

π‘₯ 𝑓(π‘₯) π‘₯ |𝑓(π‘₯)|

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4.5A – Rational Functions Part 1

A function 𝑓 is a rational function if 𝑓(π‘₯) =𝑔(π‘₯)

β„Ž(π‘₯), where 𝑔(π‘₯) and β„Ž(π‘₯) are polynomials.

The domain of 𝑓 consists of all real numbers except π‘₯ values that make the denominator equal to zero (undefined values).

Example 1 – What are the undefined values for each rational function?

a)

π‘₯+52

b)

π‘₯2π‘₯βˆ’4

c)

π‘₯2βˆ’2π‘₯βˆ’15π‘₯+4

d)

π‘₯π‘₯βˆ’72+16

Graphing Rational Functions Example 2 – Graph 𝑦 = 1

π‘₯

D: R:

An asymptote is not part of the graph. A vertical asymptote is a line the graph

approaches as the denominator approaches zero. A horizontal asymptote Is a line the graph approaches as |π‘₯| gets larger. Not every rational function has both a horizontal AND vertical asymptote.

Vertical Asymptotes are found when the denominator equals zero. Sometimes the denominator must be factored to find the vertical asymptotes (see Example 1 above).

π‘₯ 𝑦 1 2 3 10 100 1000 0 0.5 0.2 0.1 0.01 0.001 -1 -2 -10 -100 -0.5 -0.1 -0.01 -0.001

The function is not defined for π‘₯ = 0, so this is a vertical asymptote of the function, and is drawn as a dashed line. The pattern in the table can be written as:

as π‘₯ β†’ 0+ , 𝑓(π‘₯) β†’ +∞ as π‘₯ β†’ 0βˆ’ , 𝑓(π‘₯) β†’ βˆ’βˆž as π‘₯ β†’ +∞ , 𝑓(π‘₯) β†’ 0+ as π‘₯ β†’ βˆ’βˆž , 𝑓(π‘₯) β†’ 0βˆ’

The line defined by 𝑦 = 0 is said to be a horizontal asymptote of the function, and is drawn as a dashed line.

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Example 3 – Graph 𝑦 = 1

π‘₯βˆ’1

βˆ’ 1

What is the vertical asymptote? Plot it. Now, complete the table of values below:

D: R: x-int: y-int:

Another way to think about the function above is using 𝑦 =π‘₯βˆ’β„Ž1 + π‘˜, where the

intersection of the two asymptotes is (β„Ž, π‘˜). So, 𝑦 =π‘₯βˆ’11 βˆ’ 1 is just like =π‘₯1 , but shifted one unit right and one unit down.

Before we graphed 𝑦 = 1

π‘₯βˆ’1βˆ’ 1 above, we knew the vertical asymptote, but the

horizontal asymptote wasn’t apparent until after. How can we find the horizontal asymptote before?

The horizontal asymptote is the value that 𝑦 approaches as π‘₯ approaches ±∞. Consider the function (π‘₯) =𝑔(π‘₯)β„Ž(π‘₯) :

1) If β„Ž(π‘₯) is a higher power than 𝑔(π‘₯), the horizontal asymptote is 𝑦 = π‘˜. 2) If 𝑔(π‘₯) and β„Ž(π‘₯) have the same power, the horizontal asymptote is

𝑦 = π‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ π‘›π‘’π‘šπ‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ

π‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ π‘‘π‘’π‘›π‘œπ‘šπ‘–π‘›π‘Žπ‘‘π‘œπ‘Ÿ

3) If 𝑔(π‘₯) is a higher power than β„Ž(π‘₯), there is no horizontal asymptote. Let’s investigate all three cases:

a) 𝑓(π‘₯) = 1 π‘₯βˆ’2 b) 𝑓(π‘₯) = 1 π‘₯βˆ’2+ 3 c) 𝑓(π‘₯) = 2π‘₯ π‘₯2βˆ’4 d) 3π‘₯+1 π‘₯βˆ’2 e) 𝑦 = 3π‘₯(2π‘₯βˆ’1) 4π‘₯2βˆ’1 f) 𝑦 = π‘₯2 π‘₯ π‘₯ 𝑦 2 0 1 -1+ -1000 1000

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4.5B – Rational Functions Part 2 Example 1 – Graph 𝑓(π‘₯) = 1

π‘₯+3

D: R: x-int: y-int:

*To find the shape of the graph, it’s helpful to choose 𝒙 values one less and one greater than vertical asymptote(s), and also 𝒙 values approaching both horizontal and vertical asymptotes.

How would the graph be different if the function was 𝑓(π‘₯)

= βˆ’

π‘₯+31

βˆ’ 4 ?

Example 2 – Graph 𝑦 = 2π‘₯ π‘₯βˆ’1

D: R:

x-int: y-int:

π‘₯ 𝑦 π‘₯ 𝑦

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Example 3 – Graph β„Ž(π‘₯) =

2π‘₯2 π‘₯2+1 D: R: x-int: y-int: Example 4 – Graph 𝑓(π‘₯) = 2 π‘₯2βˆ’π‘₯βˆ’2 D: R: x-int: y-int: Example 5 – Graph 𝑦 =π‘₯2βˆ’3π‘₯βˆ’4 π‘₯2+2π‘₯ D: R: x-int: y-int:

*A note about horizontal asymptotes:

π‘₯ 𝑦

π‘₯ 𝑦

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Rational Functions & Graphing Part 3 – β€˜Holes’

Sometimes, due to factoring and cancellation, a rational function simplifies, which can cause cancellation of parts of the denominator, resulting in the elimination of a vertical asymptote. This will create a β€˜hole’ in the graph of the function.

Graph π’š = 𝒙+πŸ‘

π’™πŸ+π’™βˆ’πŸ”

D: Hole Coordinates: R:

Where there would have been a vertical asymptote, there is now just a β€˜hole’ in the graph. For each of the following functions, determine any vertical asymptotes and the

coordinates of any holes a) π’š = 𝒙+πŸ’ π’™πŸ+πŸ‘π’™βˆ’πŸ’

b) π’š = π’™πŸ+πŸπ’™βˆ’πŸ‘ π’™βˆ’πŸ

c) 𝒇(𝒙) = 𝒙 π’™πŸβˆ’πŸπ’™ π‘₯ 𝑦

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Graph the function & state any asymptotes, hole coordinates, domain, range, x-int, y-int 𝒇(𝒙) = βˆ’ π’™βˆ’πŸ’

π’™πŸβˆ’π’™βˆ’πŸπŸ

Figure

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