1. A parabola is the graph of the function
[tex]y=f(x)=ax^2+bx+c[/tex]
(0, 15 ) is a point of this parabola, so [tex]15=f(0)=a(0)^2+b(0)+c=c[/tex]
So c=15, which means we decrease the number of unknowns and write again:
[tex]y=f(x)=ax^2+bx+15[/tex]
2. Now, since (4, -1) is another point in the parabola:
[tex]-1=f(4)=a(4)^2+b(4)+15[/tex]
16a+4b=-16
dividing by 4:
4a+b=-4
We also know that -b/2a gives the x-coordinate of the vertex:
-b/2a=4
-b=8a
b=-8a
Substitute b=-8a in 4a+b=-4,
we get 4a-8a=-4
-4a=-4
a=1, then b=-8
So y=f(x)=x^2-8x+15
The roots of the expression, which are the x-intercepts can be found by solving the equation:
[tex]x^2-8x+15=0[/tex]
[tex]x^2-8x+16-1=0[/tex]
[tex](x-4)^2=1[/tex]
solution 1: x-4=1, x=5
solution 2: x-4=-1, x=3
The x-intercepts are (3, 0) and (5, 0)
