Question

How do you use the important points to sketch the graph of `y=2x^2+6`?

Answer 1

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Please read the explanation.

Explanation:

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We are given the quadratic equation:

`color(red)(y=f(x)=2x^2+6`

General form: `color(green)(y=f(x)=ax^2+bx+c` with `color(red)(a!=0`

Note that : `color(blue)(a=2; b=0 and c=6`

The coefficient of the `color(red)(x^2)` term is positive and hence the graph of the parabola opens upward.

Set `color(red)(x=0`, to find the y-intercept.

`rArr2(0)^2+6`

`rArr 6`

Hence, y-intecept: `color(blue)((0,6)`

Set `color(red)(y=0`, to find the x-intercept.

`2x^2+6=0`

`rArr 2x^2=-6`

`rArr x^2=-6/2=-3`, value of `color(red)(x^2` can't be negative for `color(blue)(x in RR`

Hence, there are no x-intercepts.

To find the vertex of the parabola:

`color(red)(Vertex=-b/(2a)`

`rArr -0/(2(2)=0`

Hence, the x-coordinate of the Vertex is = 0

To find the y-coordinate of the Vertex, set x = 0

`y=2(0)^2+6=6`

Hence, `color(red)(Vertex` is at: `color(blue)((0,6)`

Axis of Symmetry : `color(red)(x=0`

Graphs of `color(blue)(y=x^2` and `color(green)(y=2x^2`, will both have their axes of symmetry at `color(red)(x=0`

Graph of `color(red)(y=2x^2`, will lean closer to the y-axis and skinny

Using the above intermediate results, we can graph:

For the sake of better understanding, graphs of:

`color(red)(y=x^2, y =2x^2 and y = 2x^2+6` are all created.

Better understanding is achieved, by comparing the behavior of all the three graphs.

Hope this helps.