Question

How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry of `y=-(x+1)(x-3)`?

Answer 1

`"see explanation"`

Explanation:

`"given a quadratic in standard form"`

`•color(white)(x)y=ax^2+bx+c color(white)(x)a!=0`

`• " if "a>0" then graph opens up "uuu`

`• " if "a<0" then graph opens down "nnn`

`"here "y=-(x+1)(x-3)`

`color(white)(xxxxx)=-(x^2-2x-3)`

`color(white)(xxxxx)=-x^2+2x+3larrcolor(blue)"in standard form"`

`a<0rArr" graph opens down"`

`"the axis of symmetry is at the midpoint of the zeros and"`
`"the vertex lies on the axis of symmetry"`

`"let y = 0 to find zeros"`

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

`"equate each factor to zero and solve for x"`

`x+1=0rArrx=-1`

`x-3=0rArrx=3`

`"midpoint "=(-1+3)/2=1`

`rArrcolor(magenta)"axis of symmetry is "x=1`

`"x = 1 is also the x-coordinate of the vertex"`

`"substitute "x=1" into the equation for y"`

`y=-(2)(-2)=4`

`rArrcolor(magenta)"vertex "=(1,4)`
graph{(y+x^2-2x-3)(y-1000x+1000)((x-1)^2+(y-4)^2-0.04)=0 [-10, 10, -5, 5]}