Question

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

Answer 1

see explanation.

Explanation:

The standard form of the `color(blue)"quadratic function"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=ax^2+bx+c ; a!=0)color(white)(2/2)|)))`

`rArry=(x+2)(3x+2)=3x^2+8x+4`

`rArr"here " a=3,b=8" and " c=4`

To find the `color(blue)"x and y intercepts"`

`x=0toy=4larrcolor(red)" y-intercept"`

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

`rArrx=-2" or " x=-2/3larrcolor(red)"x-intercepts"`

To find the `color(blue)"vertex"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(x_("vertex")=-b/(2a))color(white)(2/2)|)))`

`rArrx_("vertex")=-8/6=-4/3`

Substitute this value into equation and solve for y

`rArry_("vertex")=3(-4/3)^2+8(-4/3)+4=-4/3`

`rArrcolor(red)"vertex "=(-4/3,-4/3)`

To find `color(blue)"maximum/minimum"`

`• "If " a>0" then minimum " uuu`

`• " If " a < 0" then maximum " nnn`

`"here " a=3>0" hence " uuu`
graph{3x^2+8x+4 [-11.25, 11.25, -5.63, 5.62]}