Question

Need help with this have no idea how to do it

Answer 1

Answer:

Step-by-step explanation:

x- intercept, vertex (v) and axis of symmetry, parabola form:x²+bx+c

vertex(h,k)

1- f(x)=(x+3)(x-3) change into parabola form

f(x)=x²-9  a=1,b=0 ,c=-9

h=-b/2a=0

k=f(0)=-9

vertex(0,-9) , x intercept is when f(x)=0

(x-3)(x+3)=0 either x-3=0⇒ x=3 or x+3=0 then x=-3

x=3, x=-3 (-3,0) and (3,0)

the x of symmetry is the h of the vertex=0

2-g(x)=(x+1)(x-3)

g(x)=x²-2x-3 a=1, b=-2,c=-3

h=-b/2a⇒-(-2)/2(1)⇒h=1

k=f(1)=1²-2(1)-3⇒k=-4

v(1,-4)

x of symmetry=h=1

(x+1)(x-3)=0

x+1=0⇒x=-1 (-1,0)

x-3=0 ⇒x=3  (3,0)

x intercept :-1,3

3-y=-x(x+6) ⇒y=-x²-6x a=-1,b=-6, c=0

vertex(-3,9)

x intercept:(0,0) and (-6,0)

axis of symmetry =-3

4-g(x)=2(x-5)(x-1) ⇒ 2x²-12x+10 a=2, b=-12, c=10

vertex(3,-8)

axis of symmetry=3

x intercept : (5,0), (1,0)

5) -4x(x+1)⇒-4x²-4x a=-4,b=-4

vertex(-1/2,1)

x of symmetry=-1/2

x intercept : (0,0)(-1,0)

6- f(x)=-2(x-3)² ⇒-2x²+12x-18

vertex(3,0)

x intercept (3,0)

axis of symmetry = 3