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2 years ago

Write the equation for the transformation of f(x) = |x| as shown in the graph.

Mathematics

2 answers:

OLEGan [10]2 years ago

6 0

Answer:

y = 2|x-3| + 1

Step-by-step explanation:

The vertex of g(x) = |x| is at (0,0)

We have moved the vertex to ( 3,1)

y = f(x) + C C > 0 moves it up

y = |x| + 1 for the shift up 1

y = f(x + C) C < 0 moves it right

y = |x-3| for the shift to the right 3

y = Cf(x) C > 1 stretches it in the y-direction

y = 2|x| since it is stretched by a factor of 2 in the y direction

Combining

y =2|x-3| + 1

balu736 [363]2 years ago

4 0

Answer:

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

Step-by-step explanation:

Reason why it's not +3 is because it's the reverse when graphed. But that only applies for the numbers in the brackets. If it's in the brackets and it's positive, you go to the left. If it's negative, then it goes to the right. And as you see, the V isn't on the point (0 , 3), and is in fact, above it, by 1. The graph has also been stretched vertically by 3, it looks like. Hope that was helpful for you!

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2 years ago

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2 years ago

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2 years ago

How many Solutions does this system have? (1 point)

mixas84 [53]

The given system of equation that is $2x+y=3$ and $6x=9-3y$ has infinite number of solutions.

Option -C.

Solution:

Need to determine number of solution given system of equation has.

$\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}$

Let us first bring the equation in standard form for comparison

$\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

To check how many solutions are there for system of equations $a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0$ , we need to compare ratios of $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$

In our case,

$a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3$

$a_{2} = 6, b_{2} = 3,\text{ and }c_{2} = -9$

$\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}$

As $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ , so given system of equations have infinite number of solutions.

Hence, we can conclude that system has infinite number of solutions.

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