Question 1) Function defining the table:
From the table the x-intercepts are -2 and 1. This means the factors are:
(x+2) and (x-1)
Let
[tex]h(x) = a(x + 2)(x - 1)[/tex]
The point (-1,-1) satisfy this function since it is from the same table.
[tex] - 1 = a( - 1 + 2)( - 1 - 1) \\ - 1 = - 2a \\ a = \frac{1}{2} [/tex]
Therefore the function is
[tex]h(x) = \frac{1}{2} (x + 2)(x - 1)[/tex]
We expand to get:
[tex]h(x) = \frac{1}{2} ( {x}^{2} + x - 2)[/tex]
The standard form is:
[tex]h(x) = \frac{ {x}^{2} }{2} + \frac{x}{2} - 1[/tex]
Question 3) Parabola opening up
The x-intercepts are x=3 and x=7
The factors are (x-3), (x-7)
The factored from is
[tex]y = a(x - 3)(x - 7)[/tex]
The curve passes through (5,-4)
[tex] - 4= a( 5- 3)( 5 - 7) \\ - 4= - 4a \\ a = 1[/tex]
The equation is:
[tex]y = (x - 3)(x + 7)[/tex]
Expand:
[tex]y = {x}^{2} + 7x - 3x - 21[/tex]
[tex]y = {x}^{2} + 4x - 21[/tex]
This is the standard form:
Question 3) Parabola opening down:
The x-intercepts are x=-5 and x=1
The factors are (x+5), (x-1)
The factored form is
[tex]y = - (x + 5)(x - 1)[/tex]
We expand to get:
[tex]y = - ( {x}^{2} - x + 5x - 5)[/tex]
[tex]y = - {x}^{2} - 4x + 5[/tex]
This is the standard form.
