HOW WOULd I GRAPH h(x) = 3 x SQUAREd?

Answer 1

its a parabola ...........

Answer 2

First, I always mentally substitute a "y" for the "h(x)". You're really graphing the equation y = 3x^2.

Since this equation is a second degree polynomial (i.e. the highest power of x is 2), the graph will be a parabola.

The simplest way to graph it is by just plugging in values of x, and seeing what y ends up being.

When x=0, y=3*(0*0) = 0
When x = 1, y = 3*(1*1) = 3
When x = 2, y = 3*(2*2) = 12
When x = -1, y = 3*(-1*-1) = 3
When x = -2, y = 3*(-2*-2) = 12

This gives you the points (-2,12), (-1, 3), (0,0), (1, 3) and (2, 12). Graph those points, connect the dots, and you'll see the parabola.

Hope that helps!

Answer 3

Once again using the memorized h(x) = x^2 shape of the graph - U, the 'parabola', starting at (0,0).

Now, because there is a 3 in front of the x, this tells you to "vertically stretch" it. ANY number larger than 1 tells you to stretch the graph.

Think of holding a rubber band and pulling (stretching) it from the top and bottom. This is what your graph is doing. So it looks like it is getting skinnier on the sides.

Plotting a few points will show you this as well.

Your origin will not change, the graph still starts at (0,0).

Answer 4

given a generic parabola

y(x) = a*(x-b)^2 + c

Start by plotting the graph
y(x) = x^2. Then shift the whole thing to the right by b (left if b is negative). Then shift it up by c (down if c is negative). Then stretch (or flip if a is negative) by a.

So for h(x) = 3x^2, it is still centered at (0,0) because b=0 and c=0, but is stretched by 3, so draw a regular parabola and then stretch it out vertically.

To see this, plot the points x=-1,0,1 on the parabola
y(x) = x^2 and h(x) = 3x^3

for x=-1, y(-1) = 1, h(-1) = 3
for x=0 y(0) = 0, h(0)=0
for x=1 y(1) = 1, h(1) = 3

so h goes through the same vertex as y, but its 3x higher than y.

Answer 5

its looks just like a normal parabola except its vertically stretched by 3.

when x= -2, y= 12
when x= -1, y= 3
when x= 0, y= 0
when x= 1, y=3
when x=2, y=12
when x=3, y=27
when x=4, y=48 etc..


and regual parabola goes like this
when x=-2 y=4
when x= -1 y=1
when x=0 y=0
when x=1 y=1
when x=2 y=4
when x=3 y=9
when x=4 y=16

so its like the regual only square the x value and then multiply by 3. for example. to find when x=16. find teh square of 16 which = 256. thne multiply by 3, which = 768. so when x=16 y = 768