Question

How do you graph and label the vertex and axis of symmetry of `y=-3(x-2)^2+2`?

Answer 1

vertex (2,2), axis of symmetry x=2

Explanation:

The coefficient of `x^2` is negative, signifies the parabola opens down wards.

The given equation is in vertex form, it is easily identifiable (2,2). Axis of symmetry is x=2

y intercept would be `y= -3(-2)^2+2` =-10

With these inputs the parabola can be easily graphed as shown below.

Answer 2

`"see explanation"`

Explanation:

`"to graph the parabola, would require"`

`• " coordinates of the vertex"`

`• " x and y intercepts"`

`• " shape of parabola max or min"`

`"the equation of a parabola in "color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`
where ( h , k ) are the coordinates of the vertex and a is a constant.

`y=-3(x-2)^2+2" is in this form"`

`"with " h=2,k=2`

`rArrcolor(magenta)"vertex "=(2,2)`

`"the axis of symmetry passes through the vertex"`
`"and is a vertical line with equation " x=2`

`color(blue)"Intercepts"`

`• " let x = 0, in the equation for y-intercept"`

`• " let y = 0, in the equation for x-intercepts"`

`x=0toy=-3(-2)^2+2=-10larrcolor(red)" y-intercept"`

`y=0to-3(x-2)^2+2=0`

`rArr(x-2)^2=2/3`

`color(blue)"take the square root of both sides"`

`x-2=+-sqrt(2/3)larr" note plus or minus"`

`rArrx=2+-sqrt(2/3)larr" exact values"`

`x~~2.82,x~~1.18" to 2 dec. places "larrcolor(red)" x-intercepts"`

`color(blue)"Shape of parabola"`

`• " if " a>0" then minimum " uuu`

`• " if " a<0" then maximum " nnn`

`"here " a=-3<0" hence maximum"`
graph{(y+3x^2-12x+10)(y-1000x+2000)=0 [-10, 10, -5, 5]}