Let f(x) = x^2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses

Question

Let f(x) = x^2 + 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

I'm weak in quadratic equations and I got lost trying to solve this question. Can I know what is an efficient way to solve and what goes through your mind while solving questions like these?

BrentGMATPrepNow · Accepted Answer

The given info tells us that, when x = 1 and when x = -4, the expression x² + bx + c equals 0
IMPORTANT: Notice that (x - 1)(x + 4) equals 0 when x = 1 and when x = -4
So, we can conclude that x² + bx + c is the same as (x - 1)(x + 4) 
We have: x² + bx + c = (x - 1)(x + 4)
Expand right side: x² + bx + c = x² + 3x - 1
This tells us that f(x) = x² + 3x - 4

The question asks us to determine where this function crosses the y-axis. 
This occurs at the point where x =  0 
So, plug x =  0  into the function to get: f( 0 ) =  0 ² + 3( 0 ) - 4 =   -4  
So, this function crosses the y-axis at ( 0 ,   -4  )

Answer:  A

Kurtosis · Answer

x = 0 when f(x) crosses the y axis --> f(0) = c

f(1) = 0 = 1 + b + c --> c = -b - 1
f(-4) = 0 = 16 - 4b + c --> c = 4b - 16

4b - 16 = -b - 1
b = 3

f(0) = -3 - 1 = -4

Answer: A

youcouldsailaway · Answer

f(x) = x^2 + bx + c  is of the form  ax^2 + bx + c . Thus, we can assume a = 1 in given function,  f(x) .

Now,

f(1) = 0 = 1 * 1^2 + 1 * b + c

and

f(-4) = 0 = 1 * (-4)^2 + (-4) * b + c

Now,  f(1) = 0 = f(-4)

1 + b + c = 16 - 4b + c 
 5b = 15  
 b = 3

Now, substituting b = 3 in  f(1)  above,

c = -4

In the equation  ax^2 + bx + c , c represents the y-intercept, which is the value of the y co-ordinate when x = 0. Thus, the answer is A.

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Let f(x) = x^2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses

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Let f(x) = x^2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses

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