GMAT Exam > GMAT Questions > Let f(x) = x2 + bx + c. If f(1) = 0 and f(-4)...
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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
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Let f(x) = x2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses t...
- 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) = x2 + 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 x2 + bx + c = 0, (x – a) is a factor of the equation.
(x + d)(x + e) = x2 + 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.
x2 + 3x - 4 = 0
(x - 1)(x + 4)
Solutions: f(1) = 0 or f(-4) = 0
x2 -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
x2 – x + 4x – 4 = 0
x2 + 3x – 4 = 0 - For any quadratic equation in the form ax2 + 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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Community Answer
Let f(x) = x2 + bx + c. If f(1) = 0 and f(-4) = 0, then f(x) crosses t...
- 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) = x2 + 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 x2 + bx + c = 0, (x – a) is a factor of the equation.
(x + d)(x + e) = x2 + 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.
x2 + 3x - 4 = 0
(x - 1)(x + 4)
Solutions: f(1) = 0 or f(-4) = 0
x2 -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
x2 – x + 4x – 4 = 0
x2 + 3x – 4 = 0 - For any quadratic equation in the form ax2 + 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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