TTM! 10 POINTS!

Mathematics

1 answer:

kodGreya [7K]3 years ago

8 0

Answer:

It has infinite solutions.

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The graph of Fx), shown below, resembles the graph of G(X) = x?, but it has

Katyanochek1 [597]

Answer:

$y=\left(x-a\right)^{2\ }+b$

$y=\left(x-2\right)^{2\ }+2$

Step-by-step explanation:

Since we are not given any options so we will start with the general case.

The equation of such a graph can be modeled as

$y = x^{2}$

This graph is symmetrical around the origin, but in the attached diagram, the graph is shifted in the x-axis as well as in the y-axis,

$y=\left(x-a\right)^{2\ }$

Where the variable a shifts the graph to the right of x-axis when a is positive, when it is negative it shifts the graph to the left that is negative x-axis.

$y=\left(x-a\right)^{2\ }+b$

The variable b shifts the graph upward in the positive y-axis when b is positive, when it is negative it shifts the graph to the bottom that is negative y-axis.

Since in the attached diagram, the graph is in the positive x-axis and y-axis, therefore, a and b must be positive and can be estimated as

$y=\left(x-2\right)^{2\ }+2$

The graph of the above expression is attached and it looks very similar to the given graph.