find the x-interceps and vertex of the function y=x^2+2x+6
help please!!
first factor the problem, that will give you the x intercepts.
okay so (x+3)(x+2)? i think i cant figure out how to factor it correctly
so you can't factor it correct?
i factored it buut im not sure if im right
It's not, that's because you can't factor it. The parabola does not cross the x axis
oh! so it doesnt have any x-intercepts?
right, or at least no real ones, but that's another lesson.
okay so how do you find the vertex? because dont you have to factor it first?
do you remember the quadratic formula?
yeah. do i put the equation in the quadratic formula and solve it?
not exactly, but plug in the numbers for \[\left(\begin{matrix}-b \\ 2a\end{matrix}\right)\]
this will give you the x of the vertex, (x,y)
okay so -1?
right, so plug that back into the equation for the y value
okay so the vertex is (-1,5)?
right. so x=no solution and vertex=(-1,5)
omg! thanks. can you help me with more? i suck at algebra?
sure what other questions do you have?
find the x-corrdnate of the vertex of the function below you can enter just a number. y=x^2+10+21
well can you factor it?
yeah. (x+7) (x+3)
so you can either average the x intercepts or use the equation I showed you earlier.
so the vertex is -5? so.....(-5,-4)?
or wait....it = -5?
sorry, I think I lost you for a second. Are you still here?
yeah im here
but I got the right answer for that question. now im on a different one.
yep that's right, do you still need help?
im not sure. are you leaving?
no, I'm here if you need me.
okay thanks so much!!!
no problem. hey how would I find you again if I just quickly answered another question?
um you can become my fan. and then i can message you when your online kind of like facebook
Okay, officially a fan. ask me any question any time. Except for spelling.
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