Find the x and y intercepts

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

1 year agothis one looks scary if im honest

1 year agoOkay, 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: \[y = \frac{ x^2 + 2x + 1 }{ 3x^2 + 3 }\]

1 year agookay

1 year agoi wrote that down

1 year agoUsually, 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 \(y\)

1 year agox = -1?

1 year agoHow'd you get that? Mind showing your steps?

1 year agoso 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

1 year agoand the x+1 is x =-1 for zero

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

1 year agoWhich is indeed \(x = -1\)

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

1 year agoVery good. They are both correct. Great job.

1 year agothank you

1 year ago