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Sagot :
Let’s begin by analyzing the quadratic function [tex]\( h(x) = ax^2 - 2x + c \)[/tex], where [tex]\(a\)[/tex] and [tex]\(c\)[/tex] are constants, and [tex]\(a < 0\)[/tex] because the parabola opens downward.
Given:
1. The function satisfies [tex]\( h(1) = h(-13) \)[/tex].
2. It is also given that the vertex of the parabola is [tex]\((h, k)\)[/tex], where [tex]\(k>0\)[/tex], confirming that the maximum point of the parabola is above the x-axis, as [tex]\(k\)[/tex] is a positive value.
Step 1: Determine the vertex form condition
The general form for [tex]\( h(x) = ax^2 - 2x + c \)[/tex] can be used to find the vertex. The x-coordinate [tex]\( h \)[/tex] of the vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] is given by:
[tex]\[ h = -\frac{b}{2a} \][/tex]
Here, [tex]\( b = -2 \)[/tex], so:
[tex]\[ h = -\frac{-2}{2a} = \frac{2}{2a} = \frac{1}{a} \][/tex]
Step 2: Using the known [tex]\( h(1) = h(-13) \)[/tex]
We evaluate [tex]\( h(1) \)[/tex] and [tex]\( h(-13) \)[/tex]:
[tex]\[ h(1) = a(1)^2 - 2(1) + c = a - 2 + c \][/tex]
[tex]\[ h(-13) = a(-13)^2 - 2(-13) + c = 169a + 26 + c \][/tex]
We are given:
[tex]\[ a - 2 + c = 169a + 26 + c \][/tex]
Simplifying this equation:
[tex]\[ a - 2 + c = 169a + 26 + c \][/tex]
Subtract [tex]\( c \)[/tex] from both sides:
[tex]\[ a - 2 = 169a + 26 \][/tex]
Rearranging terms to isolate [tex]\( a \)[/tex]:
[tex]\[ a - 169a = 26 + 2 \][/tex]
[tex]\[ -168a = 28 \][/tex]
[tex]\[ a = -\frac{28}{168} = -\frac{1}{6} \][/tex]
So, [tex]\( a = -\frac{1}{6} \)[/tex], which meets the condition [tex]\( -1 < a < 0 \)[/tex]. Hence, statement I is true.
Step 3: Determine the condition on [tex]\( c \)[/tex]
Given that the vertex [tex]\( (h, k) \)[/tex] has a positive [tex]\( k \)[/tex]:
[tex]\[ h = \frac{1}{a} = \frac{1}{-\frac{1}{6}} = -6 \][/tex]
The y-coordinate of the vertex [tex]\( k \)[/tex] is given by substituting [tex]\( h \)[/tex] into the function [tex]\( h(x) \)[/tex]:
[tex]\[ k = h(-6) = a(-6)^2 - 2(-6) + c = 36a + 12 + c \][/tex]
Substitute [tex]\( a = -\frac{1}{6} \)[/tex]:
[tex]\[ k = 36 \left( -\frac{1}{6} \right) + 12 + c = -6 + 12 + c = 6 + c \][/tex]
Since [tex]\( k > 0 \)[/tex]:
[tex]\[ 6 + c > 0 \][/tex]
[tex]\[ c > -6 \][/tex]
This confirms that:
[tex]\[ c > -6 \][/tex]
However, there is no direct reference to [tex]\( c \)[/tex] being greater than zero. Nevertheless, considering [tex]\( c > -6 \)[/tex] does not strictly mean [tex]\( c > 0 \)[/tex]. Therefore, statement II is not a certainty based on the given conditions alone.
Conclusion:
From the above analysis:
- I. [tex]\(-1 < a < 0\)[/tex] is true.
- II. [tex]\( c > 0 \)[/tex] does not necessarily follow from the given information.
Hence, only statement I must be true.
Given:
1. The function satisfies [tex]\( h(1) = h(-13) \)[/tex].
2. It is also given that the vertex of the parabola is [tex]\((h, k)\)[/tex], where [tex]\(k>0\)[/tex], confirming that the maximum point of the parabola is above the x-axis, as [tex]\(k\)[/tex] is a positive value.
Step 1: Determine the vertex form condition
The general form for [tex]\( h(x) = ax^2 - 2x + c \)[/tex] can be used to find the vertex. The x-coordinate [tex]\( h \)[/tex] of the vertex of a parabola [tex]\( y = ax^2 + bx + c \)[/tex] is given by:
[tex]\[ h = -\frac{b}{2a} \][/tex]
Here, [tex]\( b = -2 \)[/tex], so:
[tex]\[ h = -\frac{-2}{2a} = \frac{2}{2a} = \frac{1}{a} \][/tex]
Step 2: Using the known [tex]\( h(1) = h(-13) \)[/tex]
We evaluate [tex]\( h(1) \)[/tex] and [tex]\( h(-13) \)[/tex]:
[tex]\[ h(1) = a(1)^2 - 2(1) + c = a - 2 + c \][/tex]
[tex]\[ h(-13) = a(-13)^2 - 2(-13) + c = 169a + 26 + c \][/tex]
We are given:
[tex]\[ a - 2 + c = 169a + 26 + c \][/tex]
Simplifying this equation:
[tex]\[ a - 2 + c = 169a + 26 + c \][/tex]
Subtract [tex]\( c \)[/tex] from both sides:
[tex]\[ a - 2 = 169a + 26 \][/tex]
Rearranging terms to isolate [tex]\( a \)[/tex]:
[tex]\[ a - 169a = 26 + 2 \][/tex]
[tex]\[ -168a = 28 \][/tex]
[tex]\[ a = -\frac{28}{168} = -\frac{1}{6} \][/tex]
So, [tex]\( a = -\frac{1}{6} \)[/tex], which meets the condition [tex]\( -1 < a < 0 \)[/tex]. Hence, statement I is true.
Step 3: Determine the condition on [tex]\( c \)[/tex]
Given that the vertex [tex]\( (h, k) \)[/tex] has a positive [tex]\( k \)[/tex]:
[tex]\[ h = \frac{1}{a} = \frac{1}{-\frac{1}{6}} = -6 \][/tex]
The y-coordinate of the vertex [tex]\( k \)[/tex] is given by substituting [tex]\( h \)[/tex] into the function [tex]\( h(x) \)[/tex]:
[tex]\[ k = h(-6) = a(-6)^2 - 2(-6) + c = 36a + 12 + c \][/tex]
Substitute [tex]\( a = -\frac{1}{6} \)[/tex]:
[tex]\[ k = 36 \left( -\frac{1}{6} \right) + 12 + c = -6 + 12 + c = 6 + c \][/tex]
Since [tex]\( k > 0 \)[/tex]:
[tex]\[ 6 + c > 0 \][/tex]
[tex]\[ c > -6 \][/tex]
This confirms that:
[tex]\[ c > -6 \][/tex]
However, there is no direct reference to [tex]\( c \)[/tex] being greater than zero. Nevertheless, considering [tex]\( c > -6 \)[/tex] does not strictly mean [tex]\( c > 0 \)[/tex]. Therefore, statement II is not a certainty based on the given conditions alone.
Conclusion:
From the above analysis:
- I. [tex]\(-1 < a < 0\)[/tex] is true.
- II. [tex]\( c > 0 \)[/tex] does not necessarily follow from the given information.
Hence, only statement I must be true.
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