Question

How do you find the vertex and intercepts for `y=3x^2+5x+8`?

Answer 1

`"see explanation"`

Explanation:

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

`•color(white)(x)y=a(x-h)^2+k`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`"to obtain this form "color(blue)"complete the square"`

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

`color(white)(y)=3(x^2+2(5/6)x color(red)(+25/36)color(red)(-25/36)+8/3)`

`color(white)(y)=3(x+5/6)^2+3(-25/36+96/36)`

`color(white)(y)=3(x+5/6)^2+71/12larrcolor(blue)"in vertex form"`

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

`"to find the x-intercepts let y = 0"`

`3(x+5/6)^2+71/12=0`

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

`"this has no real solutions hence no x-intercepts"`

`"for y-intercept let x = 0"`

`y=0+0+8=8larrcolor(red)"y-intercept"`
graph{3x^2+5x+8 [-20, 20, -10, 10]}