Question

) () = 2 − 6 + 13 Translate f(x) 7 units to the left and down 5 units. Write your new function in vertex form

Answer 1

Answer:

$y= (x -4)^2 - 1$

$Vertex = (4,-1)$

Step-by-step explanation:

Given

$f(x) = x^2 + 6x+13$

Required

7 units left and 5 units down

First, we have: [7 units left]

The rule is:

$(x,y) \to (x-7,y)$

So, we have:

$f(x) = (x - 7)^2 + 6(x -7)+13$

Next, we have: [5 units down]

The rule is:

$(x,y) \to (x,y-5)$

So, we have:

$f'(x) = (x - 7)^2 + 6(x -7)+13 - 5$

$f'(x) = (x - 7)^2 + 6(x -7)+8$

Rewrite as:

$y = (x - 7)^2 + 6(x -7)+8$

Expand

$y = x^2 - 14x + 49 + 6x -42+8$

Collect like terms

$y = x^2 - 14x + 6x -42+8+ 49$

$y = x^2 -8x +15$

To write in vertex form, we have

Subtract 15 from both sides

$y -15= x^2 -8x$

Divide (-8) by 2; Add the square to both sides

$y -15+ (-8/2)^2= x^2 -8x + (-8/2)^2$

$y -15+ 16= x^2 -8x + 16$

$y +1= x^2 -8x + 16$

Expand

$y +1= x^2 -4x - 4x + 16$

Factorize

$y +1= x(x -4) -4(x - 4)$

Factor out x - 4

$y +1= (x -4)(x - 4)$

Rewrite as:

$y +1= (x -4)^2$

Make y the subject

$y= (x -4)^2 - 1$

The vertex is:

$Vertex = (4,-1)$