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Over which interval is the graph of [tex]f(x) = -x^2 + 3x + 8[/tex] increasing?

A. [tex](-\infty, 1.5)[/tex]

B. [tex](-\infty, 10.25)[/tex]

C. [tex](1.5, \infty)[/tex]

D. [tex](10.25, \infty)[/tex]


Sagot :

Sure, let's analyze the function [tex]\( f(x) = -x^2 + 3x + 8 \)[/tex] to determine the interval over which it is increasing.

1. Identifying the coefficients:
The quadratic function [tex]\( f(x) = -x^2 + 3x + 8 \)[/tex] has:
- [tex]\( a = -1 \)[/tex] (the coefficient of [tex]\( x^2 \)[/tex])
- [tex]\( b = 3 \)[/tex] (the coefficient of [tex]\( x \)[/tex])
- [tex]\( c = 8 \)[/tex] (the constant term)

2. Vertex of the quadratic function:
The vertex form of a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex] gives the x-coordinate of the vertex as [tex]\( x = \frac{-b}{2a} \)[/tex].

Substituting [tex]\( a = -1 \)[/tex] and [tex]\( b = 3 \)[/tex] into the formula, we get:
[tex]\[ x = \frac{-b}{2a} = \frac{-3}{2(-1)} = \frac{3}{2} = 1.5 \][/tex]

3. Behavior of the quadratic function:
For a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex]:
- If [tex]\( a > 0 \)[/tex] (the parabola opens upwards), the function is increasing to the right of the vertex and decreasing to the left of the vertex.
- If [tex]\( a < 0 \)[/tex] (the parabola opens downwards), the function is increasing to the left of the vertex and decreasing to the right of the vertex.

Since [tex]\( a = -1 \)[/tex] in our function, the parabola opens downwards.

4. Determining the interval of increase:
Since the parabola opens downwards, [tex]\( f(x) \)[/tex] is increasing on the interval to the left of its vertex.
Thus, [tex]\( f(x) \)[/tex] is increasing for [tex]\( x < 1.5 \)[/tex].

5. Conclusion:
- The interval where the graph of [tex]\( f(x) = -x^2 + 3x + 8 \)[/tex] is increasing is [tex]\( (-\infty, 1.5) \)[/tex].

Therefore, the correct answer is:
[tex]\[ \boxed{(-\infty, 1.5)} \][/tex]
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