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If f(x)=ax²+bx+c, for all x is f(x)<0?

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Source: — Data Sufficiency |

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by ErikaPrepScholar » Mon Dec 04, 2017 9:21 am
This problem requires us to have some foundational knowledge about quadratic equations and their graphs. A quadratic equation, like the one here, creates a parabola. This parabola can face up or down, depending on the sign of a. We can find the roots (x-intercepts) of the graph by factoring the equation or by using the quadratic formula:

$$x\ =\ \frac{-b\ \pm\sqrt{b^2-4ac}}{2a}$$

Statement 1
We should recognize this term as what comes under the square root sign in the quadratic formula. If this term is less that zero (negative), this gives us an imaginary number. This means that any possible roots (x-intercepts) of this equation are not real.

If there are no real x-intercepts on our graph, then it never intersects the x-axis. So this means our parabola can face up and have a vertex above the x-axis (giving f(x) > 0 for all x), OR our parabola can face down and have a vertex below the x-axis (giving f(x) < 0 for all x). Insufficient.

Statement 2
If a is less than zero (negative) our parabola faces down. However, we don't know where the vertex is. It could be below the x-axis (giving f(x) < 0 for all x), on the x-axis (giving f(x) ≤ 0 for all x), or above the x-axis (giving positive, zero, and negative values for f(x)). Insufficient.

Both
Combined, these statements tells us that our graph is a parabola that faces down with a vertex below the x-axis. This gives f(x) < 0 for all x. Sufficient.
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