Question

How do you graph and identify the vertex and axis of symmetry for `y=-3/5(x-4)^2+6`?

Answer 1

Vertex:`((4, 6)`,axis of symmetry:`x=4,`x intercepts:
`(0.84,0) and (7.16,0),` y intercepts: `(0, -3.6)`, additional point:(9,-9)#

Explanation:

`y= -3/5(x-4)^2+6`

This is vertex form of equation ,`y=a(x-h)^2+k ; (h,k)`

being vertex , here `h=4 ,k=6,a=-3/5= -0.6`

Since `a` is negative, parabola opens downward.

Therefore vertex is at `(h,k) or (4, 6)`

Axis of symmetry is `x= h or x = 4 ;`

y-intercept is found by putting `x=0` in the equation

`y=-3/5(x-4)^2+6= -3/5 (0-4)^2+6= -48/5+6`

or `y= -18/5= -3.6 or (0, -3.6)`

x-intercepts are found by putting `y=0` in the equation

`0=-3/5(x-4)^2+6 or 3/5(x-4)^2 =6` or

`(x-4)^2=10 or (x-4) =+- sqrt 10 :. x = 4+- sqrt 10`

`:. x~~ 0.84 , x~~ 7.16 or (0.84,0) and (7.16,0)`

Additional point:

`x= 9 , y= -3/5(9-4)^2+6=-9 or (9,-9)`

graph{-3/5(x-4)^2+6 [-20, 20, -10, 10]} [Ans]