Someone help. Foundation

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

4 0

Answer:

rotation of 90 degrees counterclockwise centered at the origin

Step-by-step explanation:

Just look at one single point of A and what must happen to it to become the corresponding point in B.

Look at point (2, 1) in figure A. Draw a segment from (0, 0) to (2, 1).

The corresponding point in figure B to (2, 1) is (-1, 2). Now draw a segment from (0, 0) to (-1, 2). Look at the relationship between the two segments.

The segment (0, 0) to (2, 1) is perpendicular to the segment (0, 0) to (-1, 2).

The transformation is a rotation of 90 degrees counterclockwise centered at the origin.

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1. Find two functions f(x) and g(x) such that f[g(x)] = x but g[f(x)] does not equal x.2. State the vertical, horizontal asympto

Harman [31]

Answer:

1. $f(x)=x^{2}$ and $g(x)=-\sqrt{x}$

2. Vertical Asymptote: x=-4

Horizontal Asymptote: y=1

Zeros: (-2,0)

Because the graph will have a hole at x=-1. This is, the graph is not defined for that particular x-value.

3. $f(x)=x^{2}+4$

It has two complex roots, so there is no place in the graph where it crosses the x-axis.

Step-by-step explanation:

1. If we find the composite function f[g(x)] with the functions:

$f(x)=x^{2}$ and g(x)=-x, we will get the following:

$f[g(x)]=(-\sqrt{x})^{2}$

which simplifies to

f[g(x)]=x

on the other hand, if we find g[f(x)] we will get the following:

$g[f(x)]=-\sqrt{x^2}$

which simplifies to:

g[f(x)]=-x

Before finding the asymptotes, it's a good idea to factor both the numerator and denominator of the function, so we get:

$f(x)=\frac{x^{2}+3x+2}{x^{2}+5x+4}$

$f(x)=\frac{(x+2)(x+1)}{(x+4)(x+1)}$

we can now see that this function can be simplified. It is important to simplify this function so we don't confuse vertical asymptotes with holes in the graph, so when simplifying we get the following:

$f(x)=\frac{x+2}{x+4}$

now we can set the denominator equal to zero.

x+4=0

and solve for x:

x=-4 this is our vertical asymptote.

For the horizontal asymptote, there is a rule that tells us that if the polynomials on the numerator and denominator have the same degree (the same greatest power), then the horizontal asymptote is found by dividing the lead coefficient of the polynomial on the numerator by the lead coefficient on the denominator. In this case the lead coefficients are 1 and 1 so the horizontal asymptote is:

$y=\frac{1}{1}$