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


See Also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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