Math / Algebra

Question4) Individua quali fra i seguenti punti non appartengono al grafico della funzione $f(x)=\frac{1-x^{2}}{x}$ [a] $(1 ; 0)$ $5(0 ; 1)$ E $\left(-2 ; \frac{3}{2}\right)$ (motiva la risposta) [] $(-1 ; 0)$ [e $\left(-3 ;-\frac{10}{3}\right)$

Studdy Solution

STEP 1

What is this asking? Which of these points don't live on the graph of the function $f(x) = \frac{1 - x^2}{x}$ ? Watch out! Make sure to plug in the x-value into the function and see if it spits out the y-value!
Don't mix them up!

STEP 2

1. Test point (1, 0)
2. Test point (0, 1)
3. Test point (-2, 3/2)
4. Test point (-1, 0)
5. Test point (-3, -10/3)

STEP 3

Alright, let's plug in $x = 1$ into our function! $f(1) = \frac{1 - (1)^2}{1} = \frac{1 - 1}{1} = \frac{0}{1} = 0$ .
Boom! This matches the y-value of 0, so the point $(1, 0)$ is on the graph.

STEP 4

Now, let's try $x = 0$ . $f(0) = \frac{1 - (0)^2}{0} = \frac{1}{0}$ .
Uh oh!
We can't divide by zero!
This means the function isn't defined at $x = 0$ , so the point $(0, 1)$ is not on the graph.

STEP 5

Let's substitute $x = -2$ . $f(-2) = \frac{1 - (-2)^2}{-2} = \frac{1 - 4}{-2} = \frac{-3}{-2} = \frac{3}{2}$ .
This matches the y-value of $\frac{\textbf{3}}{\textbf{2}}$ , so $(-2, \frac{3}{2})$ is on the graph!

STEP 6

Time for $x = -1$ ! $f(-1) = \frac{1 - (-1)^2}{-1} = \frac{1 - 1}{-1} = \frac{0}{-1} = 0$ .
This matches the y-value of 0, so $(-1, 0)$ is on the graph.

STEP 7

Lastly, let's check $x = -3$ . $f(-3) = \frac{1 - (-3)^2}{-3} = \frac{1 - 9}{-3} = \frac{-8}{-3} = \frac{8}{3}$ .
This doesn't match the y-value of $-\frac{\textbf{10}}{\textbf{3}}$ , so $(-3, -\frac{10}{3})$ is not on the graph.

STEP 8

The points that don't belong to the graph are $(0, 1)$ and $(-3, -\frac{10}{3})$ .

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Question4) Individua quali fra i seguenti punti non appartengono al grafico della funzione $f(x)=\frac{1-x^{2}}{x}$ [a] $(1 ; 0)$ $5(0 ; 1)$ E $\left(-2 ; \frac{3}{2}\right)$ (motiva la risposta) [] $(-1 ; 0)$ [e $\left(-3 ;-\frac{10}{3}\right)$

Studdy Solution

STEP 1

What is this asking? Which of these points don't live on the graph of the function $f(x) = \frac{1 - x^2}{x}$ ? Watch out! Make sure to plug in the x-value into the function and see if it spits out the y-value!
Don't mix them up!

STEP 2

1. Test point (1, 0)
2. Test point (0, 1)
3. Test point (-2, 3/2)
4. Test point (-1, 0)
5. Test point (-3, -10/3)

STEP 3

Alright, let's plug in $x = 1$ into our function! $f(1) = \frac{1 - (1)^2}{1} = \frac{1 - 1}{1} = \frac{0}{1} = 0$ .
Boom! This matches the y-value of 0, so the point $(1, 0)$ is on the graph.

STEP 4

Now, let's try $x = 0$ . $f(0) = \frac{1 - (0)^2}{0} = \frac{1}{0}$ .
Uh oh!
We can't divide by zero!
This means the function isn't defined at $x = 0$ , so the point $(0, 1)$ is not on the graph.

STEP 5

Let's substitute $x = -2$ . $f(-2) = \frac{1 - (-2)^2}{-2} = \frac{1 - 4}{-2} = \frac{-3}{-2} = \frac{3}{2}$ .
This matches the y-value of $\frac{\textbf{3}}{\textbf{2}}$ , so $(-2, \frac{3}{2})$ is on the graph!

STEP 6

Time for $x = -1$ ! $f(-1) = \frac{1 - (-1)^2}{-1} = \frac{1 - 1}{-1} = \frac{0}{-1} = 0$ .
This matches the y-value of 0, so $(-1, 0)$ is on the graph.

STEP 7

Lastly, let's check $x = -3$ . $f(-3) = \frac{1 - (-3)^2}{-3} = \frac{1 - 9}{-3} = \frac{-8}{-3} = \frac{8}{3}$ .
This doesn't match the y-value of $-\frac{\textbf{10}}{\textbf{3}}$ , so $(-3, -\frac{10}{3})$ is not on the graph.

STEP 8

The points that don't belong to the graph are $(0, 1)$ and $(-3, -\frac{10}{3})$ .

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Studdy Solution

STEP 1

What is this asking? Which of these points *don't* live on the graph of the function f(x)=1−x2xf(x) = \frac{1 - x^2}{x}f(x)=x1−x2​? Watch out! Make sure to plug in the *x*-value into the function and see if it spits out the *y*-value!Don't mix them up!

STEP 2

1. Test point (1, 0)2. Test point (0, 1)3. Test point (-2, 3/2)4. Test point (-1, 0)5. Test point (-3, -10/3)

STEP 3

Alright, let's **plug in** x=1x = 1x=1 into our function! f(1)=1−(1)21=1−11=01=0f(1) = \frac{1 - (1)^2}{1} = \frac{1 - 1}{1} = \frac{0}{1} = 0f(1)=11−(1)2​=11−1​=10​=0.Boom! This matches the *y*-value of **0**, so the point (1,0)(1, 0)(1,0) *is* on the graph.

STEP 4

Now, let's try x=0x = 0x=0. f(0)=1−(0)20=10f(0) = \frac{1 - (0)^2}{0} = \frac{1}{0}f(0)=01−(0)2​=01​.Uh oh!We can't divide by **zero**!This means the function isn't defined at x=0x = 0x=0, so the point (0,1)(0, 1)(0,1) is *not* on the graph.

STEP 5

STEP 6

Time for x=−1x = -1x=−1! f(−1)=1−(−1)2−1=1−1−1=0−1=0f(-1) = \frac{1 - (-1)^2}{-1} = \frac{1 - 1}{-1} = \frac{0}{-1} = 0f(−1)=−11−(−1)2​=−11−1​=−10​=0.This matches the *y*-value of **0**, so (−1,0)(-1, 0)(−1,0) *is* on the graph.

STEP 7

STEP 8

The points that *don't* belong to the graph are (0,1)(0, 1)(0,1) and (−3,−103)(-3, -\frac{10}{3})(−3,−310​).

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What is this asking? Which of these points don't live on the graph of the function $f(x) = \frac{1 - x^2}{x}$ ? Watch out! Make sure to plug in the x-value into the function and see if it spits out the y-value!
Don't mix them up!

1. Test point (1, 0)
2. Test point (0, 1)
3. Test point (-2, 3/2)
4. Test point (-1, 0)
5. Test point (-3, -10/3)

Alright, let's plug in $x = 1$ into our function! $f(1) = \frac{1 - (1)^2}{1} = \frac{1 - 1}{1} = \frac{0}{1} = 0$ .
Boom! This matches the y-value of 0, so the point $(1, 0)$ is on the graph.

Now, let's try $x = 0$ . $f(0) = \frac{1 - (0)^2}{0} = \frac{1}{0}$ .
Uh oh!
We can't divide by zero!
This means the function isn't defined at $x = 0$ , so the point $(0, 1)$ is not on the graph.

Time for $x = -1$ ! $f(-1) = \frac{1 - (-1)^2}{-1} = \frac{1 - 1}{-1} = \frac{0}{-1} = 0$ .
This matches the y-value of 0, so $(-1, 0)$ is on the graph.

The points that don't belong to the graph are $(0, 1)$ and $(-3, -\frac{10}{3})$ .