How do you graph #y=2x^2+6#?

1 Answer
Nallasivam V
Sep 4, 2015

Refer the explanation section

Explanation:

It is a quadratic equation.
Its vertex is at
#x=(-b)/(2a)#

'b' term is not there in the function.
Then we shall write the function like this
#ax2 + bx + c# - This is the general form of the function.
The our function shall be writter as -
#2x^2 +0x + 6#

Now find the vertex -
#x=(-b)/(2a)= (-0)/(2 xx 2)#= 0

Since #2x^2# term is positive, It must be an upward facing curve.

Take a few points on either side of 0. Calculate the y value.

x :y
-3: 24
-2: 14
-1 :8
0: 6
1 :8
2 :14
3 :24

Plot the points. Join them with the help of a smooth curve. graph{2x^2+6 [-58.5, 58.55, -29.25, 29.2]}

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