1998 AHSME Problems/Problem 14

Problem 14

A parabola has vertex of $(4,-5)$ and has two $x-$ intercepts, one positive, and one negative. If this parabola is the graph of $y = ax^2 + bx + c,$ which of $a,b,$ and $c$ must be positive?

$\mathrm{(A) \ } \text{only}\ a \qquad \mathrm{(B) \ } \text{only}\ b \qquad \mathrm{(C) \ } \text{only}\ c \qquad \mathrm{(D) \ } a\ \text{and}\ b\ \text{only} \qquad \mathrm{(E) \ } \text{none}$

Solution

The vertex of the parabola is at $(4,-5)$ . Since there are two x-intercepts, it must open upwards. If it opened downard, there would be no roots. Thus, $a > 0$ .

The x-coordinate of the vertex is $\frac{-b}{2a}$ . Since $a$ is positive, and the x-intercept is positive, the value $-b$ must be positive too, and $b$ is negative.

By Vieta, the product of the two roots is $\frac{c}{a}$ . Since the two roots are a positive number and a negative number, the product is negative. Since $a$ is positive, that means $c$ must be negative.

Thus $\boxed{A}$ is the right answer - only $a$ is positive.

Solution

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1998 AHSME Problems/Problem 14

Problem 14

Solution