1957 AHSME Problems/Problem 10

Solution

This graph generates a parabola, since the degree of $x$ is $2$ . The $x-$ coordinate of the vertex of a parabola given by $ax^2 + bx + c$ is at $\frac{-b}{2a}$ So, the vertex of this parabola is at $x = \frac{-4}{2(2)} = \frac{-4}{4} = -1$ Since the coefficient of $x^2$ is positive, at $x = -1$ , the parabola is at its minimum. Substituting $x = -1$ , we get $y = 2(-1)^2 + 4(-1) + 3 \Rrightarrow y = 2 -4 + 3 \Rrightarrow y = 1$ So our answer is $\fbox{\textbf{(C)}}$ .

~JustinLee2017

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1957 AHSME Problems/Problem 10

Solution

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