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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  [#permalink]

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New post 11 Jun 2016, 22:08
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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?
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Re: Let f(x) = x^2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses  [#permalink]

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New post 12 Jun 2016, 11:35
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whitehalo wrote:
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



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) = 0² + 3(0) - 4 = -4
So, this function crosses the y-axis at (0, -4)

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Re: Let f(x) = x^2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses  [#permalink]

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New post 11 Jun 2016, 22:49
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1
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
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Re: Let f(x) = x^2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses  [#permalink]

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Re: Let f(x) = x^2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses   [#permalink] 02 Jun 2020, 18:42

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

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