Question

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

Answer 1

Vertex `(-3/4, -73/8)`
Y-Intercept `(0,-8)`
x-intercept `((sqrt73-3)/4,0); ((-sqrt73-3)/4,0)`

Explanation:

Given -

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

`x=(-b)/(2a)=(-3)/(2xx2)=(-3)/4=-3/4`

At `x=-3/4; y=2(-3/4)^2+3(-3/4)-8`

`y=2(-9/16)-9/4-8`

`y=18/16-9/4-8=(18-36-128)/16=-146/16=-73/8`

Vertex

`(-3/4, -73/8)`

Intercept

Y-Intercept

At `x=0; y=2(0^2)+3(0)-8=-8`

`(0,-8)`

X-intercept

At `y=0;0=2x^2+3x-8`

`2x^2+3x-8=0`

`2x^2+3x=8`

`x^2+3/2x=4`

`x^2+3/2x+9/16=4+9/16=(64+9)/16=73/16`

`(x+3/4)^2=+-sqrt(73/16)`

`x+3/4=+-sqrt(73/16)=+-sqrt73/4`

`x=sqrt73/4-3/4=(sqrt73-3)/4`

`x=-sqrt73/4-3/4=(-sqrt73-3)/4`

x-intercept `((sqrt73-3)/4,0); ((-sqrt73-3)/4,0)`