How do you graph the parabola $y = 2 x^{2} + 4$ $y= 2x^2+4$ using vertex, intercepts and additional points?

Question

How do you graph the parabola $y = 2 x^{2} + 4$ $y= 2x^2+4$ using vertex, intercepts and additional points?

1 Answer

Impact of this question

8282 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-13T06:16:36

Refer explanation section

Explanation:

Given -

$y = 2 x^{2} + 4$ $y=2x^2+4$

To find the vertex, rewrite the function as

$y = 2 x^{2} + 0 x + 4$ $y=2x^2+0x+4$

x-coordinate of the vertex

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

y coordinate of the vertex

At $x = 0$ $x=0$ ; $y = 2 {(0)}^{2} + 0 (0) + 4 = 4$ $y=2(0)^2+0(0)+4=4$

Vertex $(0, 4)$ $(0, 4)$

y Intercept $(0, 4)$ $(0, 4)$

To find the x-intercept put $y = 0$ $y=0$

$2 x^{2} + 4 = 0$ $2x^2+4=0$
$2 x^{2} = - 4$ $2x^2=-4$
$x^{2} = \frac{- 4}{2}$ $x^2=(-4)/2$
$x = \pm \frac{\sqrt{- 4}}{2}$ $x=+-sqrt(-4)/2$ [The function has imaginary roots. It means, it doesn't have x-intercept.

Take a few points on either side of $x = 0$ $x=0$ . Frame a table -

enter image source here

Plot the points on a graph sheet. Join them with a smooth curve.
enter image source here

Science

Math

Humanities

... and beyond

How do you graph the parabola $y = 2 x^{2} + 4$ $y= 2x^2+4$ using vertex, intercepts and additional points?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you graph the parabola y=2x2+4y= 2x^2+4 using vertex, intercepts and additional points?

1 Answer

Explanation:

Related questions

Impact of this question

How do you graph the parabola $y = 2 x^{2} + 4$ $y= 2x^2+4$ using vertex, intercepts and additional points?