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Darina [25.2K]
3 years ago
7

A.) Reflect the parent function over the y-axis, and translate it 3 units to the right.

Mathematics
2 answers:
Salsk061 [2.6K]3 years ago
5 0
Its very hard to explain it with out a visual representation but if you put that in standard for to translate it into slope intercept form and put it on the grid the rest should be a piece of cake.
Lynna [10]3 years ago
3 0

Answer:

The correct option is D.

Step-by-step explanation:

The cubic parent function is

f(x)=\sqrt[3]{x}

The given function is

h(x)=-\sqrt[3]{x+8}

It i can be written as

h(x)=-f(x+8)                .... (1)

The function h(x) is negative of function f(x+8), therefore the graph of parent function reflected over the x-axis.

The translation is defined as

h(x)=f(x+a)+b                .... (2)

Where, a is horizontal shift and b is vertical shift.

If a>0, then the graph shifts a units left and if a<0, then the graph shifts a units right.

If b>0, then the graph shifts b units up and if b<0, then the graph shifts b units down.

From (1) and (2) we get

a=8,b=0

Since a=8>0, therefore the graph shifts 8 units left.

The graph of parent function reflected over the x-axis, and translate it 8 units to the left.

Therefore the correct option is D.

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olganol [36]

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CPCTC

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4 0
3 years ago
Part 1: What mistake did AJ make in the graph?
grigory [225]

Answer:

Part 1) AJ drawn the parabola opening upwards, instead of drawing it opening downwards

Part 2) see the explanation

Step-by-step explanation:

Part 1) What mistake did AJ make in the graph?

we have

f(x)=-(x+2)^2-1

This is the equation of a vertical parabola written in vertex form

The parabola open downward (because the leading coefficient is negative)

The vertex represent a maximum

The vertex is the point (-2,-1)

therefore

AJ drawn the parabola opening upwards, instead of drawing it opening downwards

Part 2) Evaluate any two x-values (between -5 and 5) into AJ's function. Show your work. How does your work prove that AJ made a mistake in the graph?

take the values x=-4 and x=4

For x=-4

substitute the value of x in the quadratic equation

f(x)=-(-4+2)^2-1\\f(x)=-5

For x=4

substitute the value of x in the quadratic equation

f(x)=-(4+2)^2-1\\f(x)=-37

According to AJ's graph for the value of x=-4 the function should be positive, however it is negative and for the value of x=4 the function should be positive and the function is negative

therefore

AJ made a mistake in the graph

8 0
3 years ago
Choose he axiom that allows x(10) to be written 10x
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total money = 5x+15

x=amount of weeks
7 0
3 years ago
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