REAL QUICK, SIMPLE ALGEBRA
convert f(x) into general/vertex form f(x) = x^2 + 1x -6
vertex form i know is y= a(x-h)^2 =+ k
so u can get a general idea where the x and y cordinates cross by solving for vertical and horzontal asymptotes
f(x) = x^2 + x - 6 is already in general form
i think complete the square gets u the vertex
to conver it tot the vertex form u must make it a perfect square
that is completing the square
can u show me how w the numbers its hard for me to picture this
i see where you're getting at but dont know what to plug in and solve for
that's where completing squares comes f(x) = x^2 + x - 6 = (x + 1/2)^2 -6 - 1/4 = ( x + 1/2)^2 - 25/4
so after completing the sq, ( x + 1/2)^2 - 25/4 is the answer???
yes
it's the vertex format .... but do u know how to do it > i mean completing squares ?
yeah i know compl the sq but didnt kno i could do it for this example
so by that ur vertex is (-1/2, -25/4)
thank you sooooooooooooo much @fiberdust @***[ISURU]***
u understand how to get the vertical and horzontal asymptotes this helps
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