Asked by Jake
An equation to the graph of y=x^3+3x^2+2 at its point of inflection is:
a) y=-3x+1
b)y=-3x-7
c)y=x+5
d)y=3x+1
e)y=3x+7
Please help. I am in urgent need.
a) y=-3x+1
b)y=-3x-7
c)y=x+5
d)y=3x+1
e)y=3x+7
Please help. I am in urgent need.
Answers
Answered by
MathMate
I suppose the question is:
"A <i>line tangent</i> to the graph of y=x^3+3x^2+2 at its point of inflection is"
First find the derivatives:
y' = 3x²+6x
y" = 6x+6
At the point of inflexion, y"=0, or x=-1.
Then the slope at x=-1 is
y'(-1) = 3-6=-3
There are only two choices of lines that have a slope of -3.
From these, we calculate
y(-1) = (-1)³+3(-1)²+2
=-1+3+2
=4
Which of these two lines (with slope = -3) gives y=4 at x=-1?
"A <i>line tangent</i> to the graph of y=x^3+3x^2+2 at its point of inflection is"
First find the derivatives:
y' = 3x²+6x
y" = 6x+6
At the point of inflexion, y"=0, or x=-1.
Then the slope at x=-1 is
y'(-1) = 3-6=-3
There are only two choices of lines that have a slope of -3.
From these, we calculate
y(-1) = (-1)³+3(-1)²+2
=-1+3+2
=4
Which of these two lines (with slope = -3) gives y=4 at x=-1?
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