Functions can be represented as equations, tables and graphs
The equation represents a parabola, that has a vertex of (5, -7.5)
Parabola
The equation of a parabola is represented as:
[tex]y = a(x - h)^2 + k[/tex]
Where (h,k) represents the vertex
Rewriting the equation
The equation is given as:
[tex]f(x) = \frac{5}{16}(x -1)(x - 9)[/tex]
Expand the above equation
[tex]f(x) = \frac{5}{16}(x^2 -x - 9x + 1)[/tex]
Evaluate the common factors
[tex]f(x) = \frac{5}{16}(x^2 -10x + 1)[/tex]
Expand
[tex]f(x) = \frac{5}{16}(x^2 -10x) + \frac{5}{16}[/tex]
Add and subtract k from the bracket
[tex]f(x) = \frac{5}{16}(x^2 -10x + k - k) + \frac{5}{16}[/tex]
The value of k is calculated as follows:
[tex]k = (\frac{10}{2})^2[/tex]
[tex]k = 25[/tex]
So, we have:
[tex]f(x) = \frac{5}{16}(x^2 -10x + k - k) + \frac{5}{16}[/tex]
[tex]f(x) = \frac{5}{16}(x^2 -10x + 25) -\frac{5}{16} \times 25 + \frac{5}{16}[/tex]
[tex]f(x) = \frac{5}{16}(x^2 -10x + 25) -\frac{125}{16} + \frac{5}{16}[/tex]
Take LCM
[tex]f(x) = \frac{5}{16}(x^2 -10x + 25) +\frac{-125 + 5}{16}[/tex]
[tex]f(x) = \frac{5}{16}(x^2 -10x + 25) -\frac{120}{16}[/tex]
Factorize the expression in the bracket
[tex]f(x) = \frac{5}{16}(x -5)^2 -\frac{120}{16}[/tex]
So, we plot the graph of [tex]f(x) = \frac{5}{16}(x -5)^2 -\frac{120}{16}[/tex].
See attachment for the graph
Read more about parabolas at:
https://brainly.com/question/4061870
