Question

Find the x and y intercepts

Answer 1

\[f(x) = \frac{ x^2 + 2x + 1 }{ 3x^2 + 3 }\]

Answer 2

this one looks scary if im honest

Answer 3

Okay, so to help you out, as you know the basic equation for a function on an xy plane is \(y = f(x)\) so we can just replace \(f(x)\) with \(y\) to make it less scary:

\[y = \frac{ x^2 + 2x + 1 }{ 3x^2 + 3 }\]

Answer 4

okay

Answer 5

i wrote that down

Answer 6

Usually, to find the x-intercept, you set y = 0, then solve for \(x\).
To find the y-intercept, you set x = 0, then solve for \(y\)

Answer 7

x = -1?

Answer 8

How'd you get that? Mind showing your steps?

Answer 9

so for x it would look like \[0 = \frac{ x^2 + 2x + 1 }{ 3x^2 + 3 }\]
\[0 = \frac{ (x +1)(x + 1) }{ 3(x^2 + 1) }\]
it was here where I realized I didnt need to care about the bottom
x = -1

Answer 10

and the x+1 is x =-1 for zero

Answer 11

First, never assume you don't need to count anything. But as it turns out, there is only one \(x\) intercept in this case.

Answer 12

Which is indeed \(x = -1\)

Answer 13

for y I did\[y = \frac{ 0^2 + 2(0) + 1 }{ 3(0)^2 + 3 } = \frac{ 1 }{ 3 }\]

Answer 14

Very good. They are both correct. Great job.

Answer 15

thank you