Find the x and y intercepts

Question

Taco · Answer

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

Taco · Answer

this one looks scary if im honest

Hero · Answer

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:

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

Taco · Answer

okay

Taco · Answer

i wrote that down

Hero · Answer

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 $y$

Taco · Answer

x = -1?

Hero · Answer

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

Taco · Answer

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

Taco · Answer

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

Hero · Answer

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

Hero · Answer

Which is indeed $x = -1$

Taco · Answer

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

Hero · Answer

Very good. They are both correct. Great job.

Taco · Answer

thank you