DETERMINE THE VERTEX AND THE X-INTERCEPT (or zeros) of
f(x) = (x-4)^2 - 9

Vertex: (4,-9) .      zeros: 7,1

Step-by-step explanation:

a(x-h)^2+k .   vertex is (h,k) so it's (4,-9) in this case.

(x-4)^2-9=0 .  

(x-4)(x-4)-9=0

(x^2-8x+16)-9=0

x^2-8x+7=0

(x-7)(x-1)=0

x-7=0 . x-1=0

x=7 .  x=1

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

no

step-by-step explanation:

2-7=-5

7-2=5

when you combing like terms for ech expression, 5 and -5. these numbers are not equal to each other.