document.write( "Question 1203271: Appreciate if you could explain a step by step in detail how to get the answer (C).\r
\n" ); document.write( "\n" ); document.write( "Tried to use these Y = a(x-h)² + k and -b/2a but not sure this is the way to do so.
\n" ); document.write( "Thank you!\r
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\n" ); document.write( "\n" ); document.write( "The function f is defined for all real numbers x by f(x)=ax^2+bx+c where a, b, c are constants and a is negative. In the xy-plane, the x-coordinate of the vertex of the parabola y=f(x) is -1. If t is a number for which f(t)>f(0), which of the following must be true? \r
\n" ); document.write( "\n" ); document.write( "I. -2 < t < 0
\n" ); document.write( "II. f(t) < f (-2)
\n" ); document.write( "III.f(t) > f (1)\r
\n" ); document.write( "\n" ); document.write( "(A) I only
\n" ); document.write( "(B) II only
\n" ); document.write( "(C) I and III only
\n" ); document.write( "(D) II and III only
\n" ); document.write( "(E) I, II, and III
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Algebra.Com's Answer #838703 by greenestamps(13198)\"\" \"About 
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\n" ); document.write( "The conditions tell us the parabola opens downward, with axis of symmetry at x=-1.

\n" ); document.write( "We are given that f(t) is greater than f(0).

\n" ); document.write( "By symmetry, f(-2) = f(0) (-2 and 0 are the same distance from the axis of symmetry).

\n" ); document.write( "So f(-2) = f(0), f(t) is greater than f(0), the parabola opens downward, and the axis of symmetry is at x=-1. That means that t must be between -2 and 0, which is condition I.

\n" ); document.write( "So I is true.

\n" ); document.write( "Given that f(t) is greater than f(0) and f(-2) = f(0), f(t) is also greater than f(-2). So condition II is false.

\n" ); document.write( "We are given that f(t) is greater than f(0). x=1 is farther to the right of the axis of symmetry than x=0; since the parabola opens downward, f(t) is greater than f(1).

\n" ); document.write( "So III is true.

\n" ); document.write( "I and III are true and II is false, so the answer is (C) I and III only.

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