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irina [24]
2 years ago
7

The graph shows the function f(x)= |x-h| + K. What is the value of h ?

Mathematics
1 answer:
UkoKoshka [18]2 years ago
4 0

The value of h is -1.5 if the graph shows the function f(x)= |x-h| + K or f(x) = |x + 1.5| - 3.5.

<h3>What is a function?</h3>

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

The question is incomplete.

The complete question is in the picture, please refer to the attached picture.

We have a function:

f(x) = |x - h| + k is shown in the graph.

The above function is a transformation of a parent function:

F(x) = |x|

After applying transformation we get:

f(x) = |x + 1.5| - 3.5

On comparing:

h = -1.5

k = -3.5

Thus, the value of h is -1.5 if the graph shows the function f(x)= |x-h| + K or f(x) = |x + 1.5| - 3.5.

Learn more about the function here:

brainly.com/question/5245372

#SPJ1

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Step-by-step explanation:

Original quadratic equation is 9x^{2}-3x-2=0

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x^{2} - \frac{x}{3} - \frac{2}{9}=0

Add \frac{2}{9} to both sides to get rid of the constant on the LHS

x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}  ==> x^{2} - \frac{x}{3}=\frac{2}{9}

Add \frac{1}{36}  to both sides

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}

This simplifies to

x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = \frac{1}{6}\right) we can see that

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\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}

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So the two roots are

x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}

and

x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}

7 0
2 years ago
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