Answer :
Answer:
Q1: 2nd row 2nd tile
Q2: 1st row 1st tile
Q3: 1st row 3rd tile
Q4: 2nd row 3rd tile
Step-by-step explanation:
When
[tex]y = a {(x + p)}^{2} + q[/tex]
Vertex is (-p, q)
Answer:
[tex](-1,-\frac{3}{7} )-->f(x)=\frac{1}{2}|x+1|-\frac{3}{7}\\(5,\frac{2}{3} )-->f(x)=\frac{3}{5}|x-5|+\frac{2}{3}\\(0,-\frac{4}{5} )-->f(x)=\frac{1}{2}|x|-\frac{4}{5}\\(\frac{2}{5},\frac{5}{3} )-->f(x)=\frac{3}{2}|x-\frac{2}{y5}|+\frac{5}{3}[/tex]
Step-by-step explanation:
We can solve all of them by writing the equations in a general form
[tex]f(x)=a|x+b|+c[/tex]
This is a function transformation of
[tex]f(x)=|x|[/tex]
where:
a=Factor of vertical stretch
b=Horizontal shift (- shifts right, + shifts left)
c=Vertical shift (+ shifts right, - shifts left)
The original function has its vertex in [tex](0,0)[/tex] so the horizontal shift will be the new x coordinate in the new vertex and the vertical shift will be the new y in the new vertex like this:
[tex]f(x)=\frac{1}{2}|x+1|-\frac{3}{7}\\b=1=HorizontalShiftof-1\\c=-\frac{3}{7}=VerticalShiftof-\frac{3}{7}\\(-1,-\frac{3}{7})\\f(x)=-\frac{3}{5}|x-5|+\frac{2}{3}\\b=-5=HorizontalShiftof5\\c=\frac{2}{3}=VerticalShiftof\frac{2}{3}\\(5,\frac{2}{3})\\f(x)=\frac{1}{2}|x|-\frac{4}{5}\\b=0=HorizontalShiftof0\\c=-\frac{4}{5}=VerticalShiftof\frac{4}{5}\\(0,-\frac{4}{5})\\f(x)=-\frac{3}{2}|x-\frac{2}{5}|+\frac{5}{3}\\b=-\frac{2}{5}=HorizontalShiftof\frac{2}{5}\\c=\frac{5}{3}=VerticalShiftof\frac{5}{3}\\(\frac{2}{5},\frac{5}{3})[/tex]