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

Please read the explanation.

We are given the quadratic equation:

General form:

Note that :

The coefficient of the

Set

Hence, y-intecept:

Set

Hence, there are no x-intercepts.

To find the vertex of the parabola:

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

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

Hence,

Axis of Symmetry :

Graphs of

Graph of

Using the above intermediate results, we can graph:

For the sake of better understanding, graphs of:

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

Hope this helps.

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To sketch the graph of ( y = 2x^2 + 6 ), you can identify the important points such as the vertex, y-intercept, and any x-intercepts. The vertex of the parabola is given by the point ((h, k)), where ( h = -\frac{b}{2a} ) and ( k = f(h) ) (substitute ( h ) into the equation to find ( k )). For this equation, the vertex is ((-0.75, 6.75)). The y-intercept is the point where the graph intersects the y-axis, which is at ( y = 6 ). There are no x-intercepts because the parabola opens upwards. With this information, plot the points and sketch the parabola accordingly.

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