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

Question

Hazel Anglin · Accepted Answer

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Please read the 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.

Abigail Allen · Answer

undefined 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.