Answer:
[tex]y= (x -4)^2 - 1[/tex]
[tex]Vertex = (4,-1)[/tex]
Step-by-step explanation:
Given
[tex]f(x) = x^2 + 6x+13[/tex]
Required
7 units left and 5 units down
First, we have: [7 units left]
The rule is:
[tex](x,y) \to (x-7,y)[/tex]
So, we have:
[tex]f(x) = (x - 7)^2 + 6(x -7)+13[/tex]
Next, we have: [5 units down]
The rule is:
[tex](x,y) \to (x,y-5)[/tex]
So, we have:
[tex]f'(x) = (x - 7)^2 + 6(x -7)+13 - 5[/tex]
[tex]f'(x) = (x - 7)^2 + 6(x -7)+8[/tex]
Rewrite as:
[tex]y = (x - 7)^2 + 6(x -7)+8[/tex]
Expand
[tex]y = x^2 - 14x + 49 + 6x -42+8[/tex]
Collect like terms
[tex]y = x^2 - 14x + 6x -42+8+ 49[/tex]
[tex]y = x^2 -8x +15[/tex]
To write in vertex form, we have
Subtract 15 from both sides
[tex]y -15= x^2 -8x[/tex]
Divide (-8) by 2; Add the square to both sides
[tex]y -15+ (-8/2)^2= x^2 -8x + (-8/2)^2[/tex]
[tex]y -15+ 16= x^2 -8x + 16[/tex]
[tex]y +1= x^2 -8x + 16[/tex]
Expand
[tex]y +1= x^2 -4x - 4x + 16[/tex]
Factorize
[tex]y +1= x(x -4) -4(x - 4)[/tex]
Factor out x - 4
[tex]y +1= (x -4)(x - 4)[/tex]
Rewrite as:
[tex]y +1= (x -4)^2[/tex]
Make y the subject
[tex]y= (x -4)^2 - 1[/tex]
The vertex is:
[tex]Vertex = (4,-1)[/tex]
