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

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Answer 1 · 2016-02-29T06:04:41

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

Given -

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

We shall have it as

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

Vertex

$x = \frac{- b}{2 a} = \frac{- 0}{2} \times 2 = \frac{0}{4} = 0$ $x=(-b)/(2a)=(-0)/2 xx 2=0/4=0$

At $x = 0$ $x=0$

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

Its vertex is $(0, 7)$ $(0,7)$

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

X intercept

At $y = 0$ $y=0$

$2 x^{2} - 7 = 0$ $2x^2-7=0$
$x^{2} = \frac{7}{2} = 3.5$ $x^2=7/2=3.5$
$x = \pm \sqrt{3.5}$ $x=+-sqrt(3.5$
$x = 1.87$ $x=1.87$
$x = - 1.87$ $x=-1.87$

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

How do you find the vertex and intercepts for y=2x^2-7y=2x2−7y=2x^2-7?