Question 916411: understand almost 90% of how to graph this but there are a few things I don't quiet understand.
h(x) = (2x+6)(x-3)^2(x+1)^3
y-int = (0,54)
x-int = (-3,0)(3,0)(-1,0)
My question is how do I determine what the degree is? and what the leading coefficient is?
Also, I am having confusion on what exactly a tangent is can it be thought of like a "straight line" or "the flat spot of a graph on the x-axis"
When I go to graph I understand all the parts except where there there is a flat spot for the x-intercept (-1,0) since it's an odd degree factor shouldn't it just go straight through it and up to the y-intercept? I don't understand the idea of the flattened spots when graphing.
It would be greatly appreciated if someone could explain this.
Thank you!
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! understand almost 90% of how to graph this but there are a few things I don't quiet understand.
h(x) = (2x+6)(x-3)^2(x+1)^3
y-int = (0,54)
x-int = (-3,0)(3,0)(-1,0)
My question is how do I determine what the degree is?
Multiply the highest powered term in each factor:: x*x^2*x^3 = x^6
and what the leading coefficient is?:: 2x*x^2*x^3 = 2x^6
LC = 2
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Also, I am having confusion on what exactly a tangent is can it be thought of like a "straight line" or "the flat spot of a graph on the x-axis"
Tangent to the x-axis means the graph comes to but does not pass thru the axis.
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When I go to graph I understand all the parts except where there there is a flat spot for the x-intercept (-1,0) since it's an odd degree factor shouldn't it just go straight through it and up to the y-intercept? I don't understand the idea of the flattened spots when graphing.
The graph approaches the x axis, lingers close to it, passes thru the axis,
lingers close, then moves away from the axis.
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It would be greatly appreciated if someone could explain this.
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Cheers,
Stan H.
Answer by MathLover1(20850)  (Show Source):
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