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oheneba
 3 years ago
Find the max/min values for P(x) = 2x^2 +8x1
oheneba
 3 years ago
Find the max/min values for P(x) = 2x^2 +8x1

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ludwig457
 3 years ago
Best ResponseYou've already chosen the best response.0on a graphing calculator it reads that 7 is the local maximum (y=7)

RadEn
 3 years ago
Best ResponseYou've already chosen the best response.1*The first way, find the xvertec by use formula x=b/2a = 8/2(2) = 2 to find the max value, just plug x=2 to P(x) so, P(2) = 2(2)^2+8(2)1=7 *The 2nd way, use the formula y=D/4a (with D=b^24ac) y_max = (8^24(2)(1))/4(2) = (648)/8 = 56/8 = 7

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.0@oheneba do you know calculus?

zzr0ck3r
 3 years ago
Best ResponseYou've already chosen the best response.1you dont even need calculus, just use the vertex formula you learn in algebra. you know it is a parabola that opens down, so you will have a max value. it will be the y component of the vertex
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