Grade 11 Math help po

[kriegrust]

Maam/Sir pahelp po, grade 11 math. Find x and y intercept, asymptote, and hole. Pasama po ng solution, salamat po.

 

[MARSSH]

wait po

 

[OLIVER70]

Checking Your Skills:

1. f(x) = x / (x^2 - 4)

2. g(x) = (2x + 5) / (3x + 2)

3. h(x) = (x^2 - 9) / (x + 2)



Explanation:



X-intercept: The x-intercept is the point where the graph of the function crosses the x-axis. This happens when the y-value of the function is 0. To find the x-intercept, set y to 0 and solve for x.

Y-intercept: The y-intercept is the point where the graph of the function crosses the y-axis. This happens when the x-value of the function is 0. To find the y-intercept, set x to 0 and solve for y.

Asymptote: An asymptote is a line that the graph of a function gets closer and closer to as the input value approaches a certain point, but never actually touches.

Hole: A hole in the graph of a function is a point where the function is undefined, but the graph is not continuous.

Solution:

1. f(x) = x / (x^2 - 4)

X-intercept:

0 = x / (x^2 - 4)

0 = x

x = 0

Y-intercept:

y = x / (0^2 - 4)

y = x / -4

y = 0


Asymptotes:

x^2 - 4 = 0

(x - 2)(x + 2) = 0

x = 2 or x = -2

The vertical asymptotes are at x = 2 and x = -2.

Holes: None

2. g(x) = (2x + 5) / (3x + 2)

X-intercept:

0 = (2x + 5) / (3x + 2)

0 = 2x + 5

x = -2.5

Y-intercept:

y = (2(0) + 5) / (3(0) + 2)

y = 5 / 2

Asymptotes:

3x + 2 = 0

x = -2/3

The vertical asymptote is at x = -2/3.

Holes: None

3. h(x) = (x^2 - 9) / (x + 2)

X-intercept:

0 = (x^2 - 9) / (x + 2)

0 = x^2 - 9

0 = (x - 3)(x + 3)

x = 3 or x = -3

Y-intercept:

y = (3^2 - 9) / (3 + 2)

y = 0

Asymptotes:

x + 2 = 0

x = -2

The vertical asymptote is at x = -2.

Holes: One at x = -2

To find the hole, we can factor the numerator and denominator of the function. We can see that the term (x + 2) appears in both the numerator and denominator. This means that we can cancel out these terms, which leaves us with the function f(x) = x - 3. This function is defined at x = -2, so there must be a hole at this point.


Conclusion:

Function:	X-intercept:	Y-intercept:	Asymptotes:	Holes:
f(x) = x / (x^2 - 4)	x = 0	y = 0	x = 2, x = -2	None
g(x) = (2x + 5) / (3x + 2)	x = -2.5	y = 5 / 2	x = -2/3	None
h(x) = (x^2 - 9) / (x + 2)	x = 3, x = -3	y = 0	x = -2	x = -2

 

[MARSSH]

You do not have permission to view the full content of this post. Log in or register now.

You do not have permission to view the full content of this post. Log in or register now.

 

[kriegrust]

Checking Your Skills:

1. f(x) = x / (x^2 - 4)

2. g(x) = (2x + 5) / (3x + 2)

3. h(x) = (x^2 - 9) / (x + 2)



Explanation:



X-intercept: The x-intercept is the point where the graph of the function crosses the x-axis. This happens when the y-value of the function is 0. To find the x-intercept, set y to 0 and solve for x.

Y-intercept: The y-intercept is the point where the graph of the function crosses the y-axis. This happens when the x-value of the function is 0. To find the y-intercept, set x to 0 and solve for y.

Asymptote: An asymptote is a line that the graph of a function gets closer and closer to as the input value approaches a certain point, but never actually touches.

Hole: A hole in the graph of a function is a point where the function is undefined, but the graph is not continuous.

Solution:

1. f(x) = x / (x^2 - 4)

X-intercept:

0 = x / (x^2 - 4)

0 = x

x = 0

Y-intercept:

y = x / (0^2 - 4)

y = x / -4

y = 0


Asymptotes:

x^2 - 4 = 0

(x - 2)(x + 2) = 0

x = 2 or x = -2

The vertical asymptotes are at x = 2 and x = -2.

Holes: None

2. g(x) = (2x + 5) / (3x + 2)

X-intercept:

0 = (2x + 5) / (3x + 2)

0 = 2x + 5

x = -2.5

Y-intercept:

y = (2(0) + 5) / (3(0) + 2)

y = 5 / 2

Asymptotes:

3x + 2 = 0

x = -2/3

The vertical asymptote is at x = -2/3.

Holes: None

3. h(x) = (x^2 - 9) / (x + 2)

X-intercept:

0 = (x^2 - 9) / (x + 2)

0 = x^2 - 9

0 = (x - 3)(x + 3)

x = 3 or x = -3

Y-intercept:

y = (3^2 - 9) / (3 + 2)

y = 0

Asymptotes:

x + 2 = 0

x = -2

The vertical asymptote is at x = -2.

Holes: One at x = -2

To find the hole, we can factor the numerator and denominator of the function. We can see that the term (x + 2) appears in both the numerator and denominator. This means that we can cancel out these terms, which leaves us with the function f(x) = x - 3. This function is defined at x = -2, so there must be a hole at this point.


Conclusion:

Function:	X-intercept:	Y-intercept:	Asymptotes:	Holes:
f(x) = x / (x^2 - 4)	x = 0	y = 0	x = 2, x = -2	None
g(x) = (2x + 5) / (3x + 2)	x = -2.5	y = 5 / 2	x = -2/3	None
h(x) = (x^2 - 9) / (x + 2)	x = 3, x = -3	y = 0	x = -2	x = -2

Boss tanong lang po sa number 3, yung na substitute niyo po sa x sa may y intercept
is 3. Hindi po ba dapat 0 rin?