Question

How do you find the vertex of the parabola `y=3x^2+5x+8`?

Answer 1

`(-5/6, 71/12)`

Explanation:

Given parabola:

`y=3x^2+5x+8`

`y=3(x^2+5/3 x)+8`

`y=3(x^2+2(5/6) x+(5/6)^2)-3(5/6)^2+8`

`y=3(x+5/6)^2-\frac{25}{12}+8`

`y=3(x+5/6)^2+\frac{71}{12}`

`3(x+5/6)^2=y-71/12`

`(x+5/6)^2=1/3(y-71/12)`

The above equation is in standard formula of upward parabola: `(x-x_1)^2=4a(y-y_1)\ \forall \ \ a>0` which has

Vertex: `(x-x_1=0, y-y_1=0)`

`(x+5/6=0, y-71/12=0)`

`\equiv(-5/6, 71/12)`

Answer 2

`"vertex "=(-5/6,71/12)`

Explanation:

`"given a parabola in "color(blue)"standard form"`

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

`"then the x-coordinate of the vertex is"`

`•color(white)(x)x_(color(red)"vertex")=-b/(2a)`

`y=3x^2+5x+8" is in standard form"`

`"with "a=3,b=5" and "c=8`

`x_("vertex")=-5/6`

`"substitute this value into the equation for y-coordinate"`

`y_("vertex")=3(-5/6)^2+5(-5/6)+8`

`color(white)(xxxx)=25/12-50/12+96/12=71/12`

`color(magenta)"vertex "=(-5/6,71/12)`