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Mathematics 15 Online
OpenStudy (anonymous):

Tell whether y=-3(x+1)^2+4 has a minimum or maximum value. then find that value.

OpenStudy (tamtoan):

negative coefficient, what do you think? parabola up or down?

OpenStudy (anonymous):

differentiate this eqn and equate it to zero to find the x value substitute the x value to find whether your getting minimum or maximum value

OpenStudy (anonymous):

uhmm.isnt it a maximum ? i just dont know how to find the value.

OpenStudy (tamtoan):

maximum is correct, then use the method chandhuru to get the value :) if you haven't learn calculus, i think there are formula for finding the vertex isn't it?, i just don't remember what it is :)

OpenStudy (anonymous):

no im not in calculus lol. im taking it next year. i think the formula is -b/2(a) . i got 1/2 but im not sure if thats correct

OpenStudy (tamtoan):

the value should be a point, it has (x,y) coordinate :)

OpenStudy (anonymous):

so what i did was wrong ? >_< .

OpenStudy (anonymous):

wait so , is the x value 1/2 ? but i just need to plug it in to get y?

OpenStudy (tamtoan):

what is b and what is 2a in this function?

OpenStudy (tamtoan):

or equation? if you set y = 0

OpenStudy (anonymous):

well form y=-3(x+1)^2+4 = -3x^2-3+4 so b is -3 & a is -3 right ?

OpenStudy (anonymous):

from*

OpenStudy (tamtoan):

no...expansion of (x+1)^2 = (x+1)(x+1) = x^2 + 2x + 1, try again :)

OpenStudy (anonymous):

what about the -3? where does that fit in ? >_<

OpenStudy (tamtoan):

after expand that, you distribute the -3 in...then simply...only until you have the equation in the form...ax^2 + bx + c = 0 ...then you can correctly know the value of a,b and c

OpenStudy (tamtoan):

simply = simplify

OpenStudy (anonymous):

wait ... so it will be -3(x+1)(x+1)+4 ?

OpenStudy (tamtoan):

no, what is (x +1)(x+1) equal to ?

OpenStudy (anonymous):

x^2+2x+1

OpenStudy (tamtoan):

ok then what is -3(x^2 + 2x +1) equal to ? distribute the -3 inside

OpenStudy (anonymous):

-3x^2-6x-3 ?

OpenStudy (tamtoan):

yes then you also have +4 ..rewrite the whole equation with +4

OpenStudy (anonymous):

so -3x^2-6x-1

OpenStudy (tamtoan):

-3 +4 = 1 :) it's not -1...also after that, your equation is -3x^2 - 6x +1 = 0...now what is your a and b and c ?

OpenStudy (anonymous):

oh yea lmfao . so a is -3 b is -6 & c is 1

OpenStudy (tamtoan):

:) then using formula to find x and y coordinate of the vertex

OpenStudy (anonymous):

so i plugged it into the eqtn . -b/2a & the answer is -1 ?

OpenStudy (anonymous):

well for x.

OpenStudy (tamtoan):

yeah, you got x, now sub in and find y...then you have the value (the point) of the max.

OpenStudy (anonymous):

okay thanks !!

OpenStudy (tamtoan):

np :) enjoy and have a good weekend

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