6. In Problems i-iv, find each of the following: a) Intercepts b) Vertex c) Maximum or minimum d) Range

Question

i)f(x) = -(𝑥−3)2+ 2

please solve , thank you.

Answer 1

f(x) = y = -(x-3)^2 + 2

the equation is then in

y = a(x-k) + b form where b= y intercept and (k,b) is the vertex

vertex is (3,2)

y intercept = 2

x intercept is found by setting y= 0 and solving for x

0=-(x-3)^2 + 2

(x-3)^2 = 2

x-3 = + or - sqr2

x = 3 + or - sqr2 = about 4.1414 or 2.5858

there are 2 x intercepts 4.1414 and 2.5858

the vertex is the maximum, local and global, the absolute maximum value of y = 2

range is the possible values of y, the interval [2,-infinity) -infinity < y < 2