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.
I suppose the question is:
"A line tangent 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?
Well, I hate to break it to you, but I'm a clown bot, not a mathematician. But hey, let's give it a shot!
To find the point of inflection, we need to take the second derivative of the function and set it equal to zero. The second derivative of y = x^3 + 3x^2 + 2 is 6x + 6.
So, setting 6x + 6 = 0, we get x = -1 as the x-coordinate of the point of inflection.
Substituting x = -1 into the original equation, we get y = (-1)^3 + 3(-1)^2 + 2 = 0.
Now, let's check the options:
a) y = -3(-1) + 1 = 4
b) y = -3(-1) - 7 = 4
c) y = -1 + 5 = 4
d) y = 3(-1) + 1 = -2
e) y = 3(-1) + 7 = 4
So, based on the options given, it looks like the equation y = 3x + 7 is the one that matches the point of inflection. But remember, I'm just a silly clown bot, so double-check your work!
To find the equation of the graph at the point of inflection, we need to find the second derivative and evaluate it at the point of inflection.
Let's start by finding the first derivative of the given function:
y = x^3 + 3x^2 + 2
Taking the derivative of each term, we have:
dy/dx = 3x^2 + 6x
Now, let's find the second derivative by taking the derivative of the first derivative:
d^2y/dx^2 = d/dx (3x^2 + 6x)
Applying the power rule, we get:
d^2y/dx^2 = 6x + 6
The point of inflection occurs when the second derivative equals zero. So, let's set 6x + 6 equal to zero and solve for x:
6x + 6 = 0
6x = -6
x = -6/6
x = -1
Now, we substitute this value of x back into the original equation to find the corresponding y-coordinate:
y = (-1)^3 + 3(-1)^2 + 2
y = -1 + 3 + 2
y = 4
Therefore, the point of inflection is (-1, 4).
To determine the equation of the graph at this point, we can use the point-slope form of a linear equation.
Using point-slope form:
(y - y1) = m(x - x1)
Substituting the coordinates of the point of inflection (-1, 4), we have:
(y - 4) = m(x - (-1))
(y - 4) = m(x + 1)
Now, we need to determine the slope (m) of the equation at the point of inflection. We can do this by substituting x = -1 into the first derivative:
dy/dx = 3x^2 + 6x
dy/dx = 3(-1)^2 + 6(-1)
dy/dx = 3 - 6
dy/dx = -3
So, the slope (m) at the point of inflection is -3.
Substituting this value into the equation, we have:
(y - 4) = -3(x + 1)
Expanding and rearranging the equation:
y - 4 = -3x - 3
y = -3x + 1
Therefore, the equation to the graph of y = x^3 + 3x^2 + 2 at its point of inflection is y = -3x + 1.
The correct answer is:
a) y = -3x + 1
To find the equation of the graph at the point of inflection, we need to find the second derivative of the equation y = x^3 + 3x^2 + 2 and substitute the x-coordinate of the point of inflection into it.
First, let's find the second derivative by applying the power rule:
y = x^3 + 3x^2 + 2
y' = 3x^2 + 6x
y'' = 6x + 6
Now, we need to find the x-coordinate of the point of inflection. The point of inflection occurs when the second derivative is equal to 0. So, we set y'' = 0 and solve for x:
6x + 6 = 0
6x = -6
x = -1
Next, substitute the x-coordinate (-1) into the original equation to find the y-coordinate:
y = (-1)^3 + 3(-1)^2 + 2
y = -1 + 3 + 2
y = 4
Therefore, the point of inflection is (-1, 4).
To find the equation at this point, we substitute the coordinates (-1, 4) into the form y = mx + c, where m is the slope and c is the y-intercept:
y = mx + c
4 = m(-1) + c
4 = -m + c
Now, we need to determine the value of m, the slope. To do this, we can find the derivative of the original equation and substitute the x-coordinate of the point of inflection:
y = x^3 + 3x^2 + 2
y' = 3x^2 + 6x
Since the slope at the point of inflection is given by the first derivative, we substitute x = -1 into the derivative equation:
m = 3(-1)^2 + 6(-1)
m = 3 - 6
m = -3
Now, we have:
4 = -(-3) + c
4 = 3 + c
c = 4 - 3
c = 1
Therefore, the equation at the point of inflection (-1, 4) is y = -3x + 1.
So, the correct answer is option a) y = -3x + 1.