# Thread: Finding X intercepts in a parabola

1. ## Finding X intercepts in a parabola

Hi all! I found myself needing some desperate help on a homework problem of mine, so I found this site on google, I can already say i can see myself on here frequently lol.

I have homework that wants me to find the vertex, and x and y intercepts' of a parabola. I can very easily find the vertex and the y intercepts with no problem, but the X intercept is giving me some trouble.

The problem is f(x)=-3X^2-4X+5

My professor has tought me to factor the problem and it has worked on a previous problem, but it doesn't work on this one. The back of the book says the X intercepts are -2.12 and .79 but no clue how to get that. Hope you guys can help

Edit: sorry, I just realized I was probably supposed to post this in the Math Challenging Problems forum, my apologies

2. The x-intercepts of $y = ax^2+bx+c$ can be found via the formula

$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In your case $a=-3, b=-4$ and $c= 5$

Have a go!

3. Ty so much for the reply I got it now.

4. you could also write it in the form:

$ax^2 + bx + c = p(x+q)^2+r$

this will give the vertex (-q,r), and p will tell you whether its a max or min.

setting it equal to 0 will solve for x

the book answers are approximations only