Respuesta :
Given:
The given function is:
[tex]y=-32x^2+90x+3[/tex]
To find:
The range of the given function.
Solution:
We have,
[tex]y=-32x^2+90x+3[/tex]
It is a quadratic function because the highest power of the variable x is 2.
Here, the leading coefficient is -32 which is negative. So, the graph of the given function is a downward parabola.
If a quadratic function is [tex]f(x)=ax^2+bx+c[/tex], then the vertex of the quadratic function is:
[tex]Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)[/tex]
In the given function, [tex]a=-32,b=90,c=3[/tex].
[tex]\dfrac{-b}{2a}=\dfrac{-90}{2(-32)}[/tex]
[tex]\dfrac{-b}{2a}=\dfrac{-90}{-64}[/tex]
[tex]\dfrac{-b}{2a}=\dfrac{45}{32}[/tex]
The value of the given function at [tex]x=\dfrac{45}{32}[/tex] is:
[tex]y=-32(\dfrac{45}{32})^2+90(\dfrac{45}{32})+3[/tex]
[tex]y=\dfrac{2121}{32}[/tex]
The vertex of the given downward parabola is [tex]\left(\dfrac{45}{32},\dfrac{2121}{32}\right)[/tex]. It means the maximum value of the function is [tex]y=\dfrac{2121}{32}[/tex]. So,
[tex]Range=\left\{y|y\leq \dfrac{2121}{32}\right\}[/tex]
[tex]Range=\left(-\infty, \dfrac{2121}{32}\right ][/tex]
Therefore, the range of the given function is [tex]\left (-\infty, \dfrac{2121}{32}\right ][/tex].
