How do you find the vertex and intercepts for $y=2x^2-7$ ?

Question

3331 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-29T06:04:41

Its vertex is $(0,7)$
Its Y- intercept is also $(0,7)$
Its X-intercepts are $(1.87, 0); (-1.87,0)$

Given -

$y=y=2x^2-7$

We shall have it as

$y=2x^2+0x-7$

Vertex

$x=(-b)/(2a)=(-0)/2 xx 2=0/4=0$

At $x=0$

$y=2(0)^2-7=7$

Its vertex is $(0,7)$

Its Y- intercept is also $(0,7)$

X intercept

At $y=0$

$2x^2-7=0$
$x^2=7/2=3.5$
$x=+-sqrt(3.5$
$x=1.87$
$x=-1.87$

Its X-intercepts are $(1.87, 0); (-1.87,0)$

How do you find the vertex and intercepts for y=2x^2-7y=2x^2-7?