Quadratic function [tex]h[/tex] is defined by [tex]h(x)=a x^2-2 x+c[/tex], where [tex]a[/tex] and [tex]c[/tex] are constants. The graph of [tex]y = h(x)[/tex] in the [tex]xy[/tex]-plane is a parabola that opens downward and has a vertex of [tex](h, k)[/tex], where [tex]k \ \textgreater \ 0[/tex]. If [tex]h(1) = h(-13)[/tex], which of the following must be true?

I. [tex]-1 \ \textless \ a \ \textless \ 0[/tex]
II. [tex]c \ \textgreater \ 0[/tex]



Answer :

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.