Tell whether y=-3(x+1)^2+4 has a minimum or maximum value. then find that value.
negative coefficient, what do you think? parabola up or down?
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
uhmm.isnt it a maximum ? i just dont know how to find the value.
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 :)
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
the value should be a point, it has (x,y) coordinate :)
so what i did was wrong ? >_< .
wait so , is the x value 1/2 ? but i just need to plug it in to get y?
what is b and what is 2a in this function?
or equation? if you set y = 0
well form y=-3(x+1)^2+4 = -3x^2-3+4 so b is -3 & a is -3 right ?
from*
no...expansion of (x+1)^2 = (x+1)(x+1) = x^2 + 2x + 1, try again :)
what about the -3? where does that fit in ? >_<
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
simply = simplify
wait ... so it will be -3(x+1)(x+1)+4 ?
no, what is (x +1)(x+1) equal to ?
x^2+2x+1
ok then what is -3(x^2 + 2x +1) equal to ? distribute the -3 inside
-3x^2-6x-3 ?
yes then you also have +4 ..rewrite the whole equation with +4
so -3x^2-6x-1
-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 ?
oh yea lmfao . so a is -3 b is -6 & c is 1
:) then using formula to find x and y coordinate of the vertex
so i plugged it into the eqtn . -b/2a & the answer is -1 ?
well for x.
yeah, you got x, now sub in and find y...then you have the value (the point) of the max.
okay thanks !!
np :) enjoy and have a good weekend
Join our real-time social learning platform and learn together with your friends!