# Practice GMAT Problem Solving Question

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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?Correct Answer:

**A**- This problem requires working backward from solutions to a quadratic equation to the equation itself.

Instead of factoring a quadratic equation to solve it (the process many questions require), this question requires you to go the other way—from solutions to factors.

f(x) = x^{2}+ bx + c = 0 - The key to unlocking this problem is recognizing that if f(a) = 0 and f(x) is a quadratic equation in the form x
^{2}+ bx + c = 0, (x – a) is a factor of the equation.

(x + d)(x + e) = x^{2}+ dex + de = 0 - If this past step does not make sense immediately, take a look at a few examples and allow them to convince you of this truth.

x^{2}+ 3x - 4 = 0

(x - 1)(x+4)

Solutions: f(1) = 0 or f(-4) = 0

x^{2}-9x + 20 = 0

(x – 4)(x – 5)

Solutions: f(4) = 0 or f(5) = 0 - Following this same principle, you know that since f(1) = 0 and f(-4) = 0, (x – 1) and (x + 4) are factors.

(x – 1)(x + 4) = 0 - Multiply these factors together to form a quadratic equation.

(x – 1)(x + 4) = 0

x^{2}– x + 4x – 4 = 0

x^{2}+ 3x – 4 = 0 - For any quadratic equation in the form ax
^{2}+ bx + c = 0, the y-intercept (i.e., the place where the equation crosses the y-axis) is c.

Even if you did not know this, you could still find the y-axis. The line will cross the y-axis where x = 0.

f(0) = (0)^{2}+ 3(0) – 4 = -4 - Y-Intercept is -4

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