Question

How do you graph `y=(x-2)^2-3`?

Answer 1

see explanation.

Explanation:

To sketch the parabola we require.

`• " x and y intercepts"`

`• " coordinates of vertex"`

`• " whether maximum or minimum"`

`color(blue)"x and y intercepts"`

`"let x = 0"toy=(-2)^2-3=1to(0,1)`

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

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

Take `color(blue)"square root of both sides"`

`sqrt((x-2)^2)=+-sqrt3`

`rArrx-2=+-sqrt3`

`rArrx=sqrt3+2~~3.73to(3.73,0)`

`rArrx=sqrt3-2~~0.27to(0.27,0)`

`color(blue)"coordinates of vertex"`

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

`rArr(2,-3)larrcolor(red)" coordinates of vertex"`

`color(blue)"maximum/minimum"`

`• " If " a>0" then minimum " uuu`

`• " If " a>0" then maximum " nnn`

`"here " a>0rArr" minimum"`

Utilising these key elements should enable a sketch of the graph to be made.
graph{(x-2)^2-3 [-10, 10, -5, 5]}

Answer 2

`"graph of the function given has been shown in the fallowing figure."`

Explanation:

`y=(x-2)^2-3`

`y=x^2-4x+4-3`

`y=x^2-4x+1" "(1)`

`"we write x=0 in the equation above (1) to find y-intercept"`

`y=0^2-4*0+1" , "y=1 " , "D(0,1)`

`"we write y=0 in the equation above (1) to find x-intercept"`

`0=x^2-4x+1`

`x_1=(-b-sqrt(b^2-4ac))/(2a)`

`x_1=(4-sqrt(16-4*1*1))/(2*1)=(4-sqrt12)/2=0.27`

`x_1=(-b+sqrt(b^2-4ac))/(2a)`

`x_1=(4+sqrt(16-4*1*1))/(2*1)=(4+sqrt12)/2=3.73`

`B(0.27,0)" , "C(3.73,0)`

`"now let us find vertex "`

`"Let's equal zero by taking the derivative of the original equation."`

`d/(d x)(x^2-4x+1)=0`

`2x-4=0`

`2x=4" , "x=4/2=2`

`"let us put x=2 in the original function "`

`y=2^2-4*2+1`

`y=4-8+1`

`y=-3`

`D(2,-3)`