Click on all that are true about the graph of y > x^2 + 4x + 4. The parabola opens up. The line of symmetry is y = -2. The vertex is at (-2, 0). The parabola is dashed. The parabola is shaded outside.
@satellite73 @jim_thompson5910
how far did you get?
Umm... The first one and the last two..?
are you able to graph y = x^2 + 4x + 4 ?
I don't know... I don't have a graph
do you have a graphing calculator?
No...
ok, here is what the graph looks like http://fooplot.com/plot/83f01cmra4
you can use that website to graph other functions as well
so where is the vertex on that blue parabolic curve?
Okay I don't understand...
are you able to see the graph/picture in that link I sent?
Yes
what does it look like?
It opens up, it is on -2, it is solid, but I don't know where it is shaded at
where is the lowest point as an ordered pair?
What?
for example, (-1,1) is on the graph but it is not the lowest point on this curve
do you see how (-1,1) is a point on the parabola?
No. I don't even know what a parabola is... I ask all my questions on here lol
Is it the first three answers?
a parabola is that bowl-shaped curve
Alright
the lowest point (when the parabola opens upward) is the vertex
what is the vertex of this parabola?
-2 ?
it has to be an ordered pair
in the form (x,y)
(-2, 0) ?
much better
you start at (0,0) and you go 2 units to the left since x =-2 and you don't move along the y axis since y = 0
The line of symmetry is y = -2. that is false because the axis of symmetry is really x = -2
The statement "The line of symmetry is y = -2." would only be true if the parabola opened up sideways, but it does not.
So that covers the equation y = x^2 + 4x + 4 to graph y > x^2 + 4x + 4, we simply graph y = x^2 + 4x + 4, then we follow these steps step 1) make the parabola a dashed line (because we do NOT include the boundary) step 2) shade above the parabola (ie on the inside of it)
so "The parabola is dashed." is true but "The parabola is shaded outside." is false
So it is the first, third, and fourth answers?
That is correct.
Thank you:) I have some other questions to post. Would you mind looking at them?
I'll help with 2 more
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