Coordinate geometry parabola

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Coordinate geometry parabola

by vipulgoyal » Fri Jul 03, 2015 1:33 am
If the graph represented by p(x) = ax^2 +bx +c passes through (-1,0) and (3,0) , is p(5) >0 ?
1. The y intercept of graph is 3
2. P(-4) < p(-3)

OA not given

Source- jamboree

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by theCEO » Fri Jul 03, 2015 8:55 pm
If a line passes through point A, then point A satisfies the equation y = ax^2 +bx +c

Point(-1,0)
Substituting x = -1 and y = 0 in the equation y = ax^2 +bx +c, we get a-b-c = 0

Point(3,0)
Substituting x = 3 and y = 0 in the equation y = ax^2 +bx +c, we get 9a+3b-c = 0

9a+3b-c = a-b-c = 0
8a+4b = 0
2a+1b = 0
b = =-2a
y = ax^2 +bx +c = ax^2 -2ax +c

1)
The y intercept of graph is 3
y intecept occurs when x=0,
y = ax^2 -2ax +c = 0 + 0 + 3; c=3
y = ax^2 -2ax +3
a = (y-3)/(x^2 - 2x)

plugging in Point(-1,0)
a = -3/-1= 3

equation = y = 3x^2 - 6x + 3
if we replace X with 5, we can tell if p(5)>0
therefore statement is sufficent

2)
y = ax^2 -2ax +c
P(-4) = 16a+8a+c = 24a+c
P(-3) = 9a+6a+c = 15a+c
P(5) = 25a-10a+c = 15a+c

If P(-4) < p(-3)
24a+c < 15a+c
24a < 15a
9a < 0
a <0; a is negative
if a is -0.01 -> P(5)= -0.15+c
if a is -10 -> P(5)= -150+c
we dont know the value of c
Insufficent

ans = a