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Rooted around the Y axis

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by MBA.Aspirant » Tue Jul 19, 2011 10:24 am
inertia2010 wrote:Please help:

Let f(x) = x2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses the y-axis at what y-coordinate?
A) -4
B) -1
C) 0
D) 1
E) 4

Thanks!!
f(x) = x2 + bx + c

c is the y-intercept


f(1) = 0

0 = 1+b+c

c= -1-b

f(-4) = 0

0 = 16 -4b +c

0 = 16 - 4b -1-b

5b= 15

b = 3

c = -1 -3 = -4

so y-intercept is -4
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by winniethepooh » Tue Jul 19, 2011 11:24 am
The answer is -4.

x^2+bx+c
f(1)= 1^2+1b+c= 0
___ = b+c = -1 ____(1)

f(-4)= 16-4b+c=0
____ = -4b+c = -16____(2)

Equating equations (1) & (2)

b= 3

substituting in equation(1)
c= -4 ____ (3)

At y axis x co-ordinate will be 0.
SO, f(0)= x^2+bx+c = 0^2+0b-4 = -4.{from equation (3)}

Hence, A.
Last edited by winniethepooh on Tue Jul 19, 2011 8:33 pm, edited 1 time in total.
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by Anurag@Gurome » Tue Jul 19, 2011 7:58 pm
inertia2010 wrote:Please help:

Let f(x) = x2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses the y-axis at what y-coordinate?
A) -4
B) -1
C) 0
D) 1
E) 4

Thanks!!
Let us take f(x) = y.
Then y = x² + bx + c
y(at x = 1) = 1 + b + c = 0...Equation 1
y(at x = -4) = 16 - 4b + c = 0...Equation 2
Equation 1: -c = 1 + b
Equation 2: -c = 16 -4b
Equating, we get 1 + b = 16 - 4b or b = 3, which implies c = -4
Now on y-axis, x = 0, so y (at x = 0) = c = -4

The correct answer is A.
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by winniethepooh » Tue Jul 19, 2011 8:35 pm
Err. Sorry for the error.Made the changes!
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