select the equations of all lines through the orgin that are tangent to the curve y=x^3-8x^2-7x
y'=3x^2-16x-7
y=mx+b
y=(3x^2-16x-7)x+b on the line.
b=o, as it goes through origin.
y=x^3-8x^2-7x on the curve,
so, where are x,y the same on the line and curve.
Y line=ycurve
3x^3-16x^2-7x=x^3-8x^2-7x
2x^3-8x=0
2x(x^2-4)=0
x=0, x=+-2
so you have these points for x, figure y at those points and you have it.
To find the equations of lines through the origin that are tangent to the curve y = x^3 - 8x^2 - 7x, we need to find the points where the curve and the lines intersect.
Step 1: Find the derivative of the curve
The derivative of the equation y = x^3 - 8x^2 - 7x is y' = 3x^2 - 16x - 7.
Step 2: Set the derivative equal to the slope of the line passing through the origin
Since the line passes through the origin, its slope is given by y/x. So, we set y' equal to y/x and solve for x.
3x^2 - 16x - 7 = (x^3 - 8x^2 - 7x) / x
Step 3: Simplify the equation
Multiply both sides of the equation by x to get rid of the fraction:
3x^3 - 16x^2 - 7x = x^3 - 8x^2 - 7x
3x^3 - 16x^2 - 7x - x^3 + 8x^2 + 7x = 0
2x^3 - 8x^2 = 0
Step 4: Factor out x^2
2x^2 (x - 4) = 0
Step 5: Solve for x
Set each factor equal to zero and solve for x:
2x^2 = 0
x = 0
x - 4 = 0
x = 4
Step 6: Find the y-values
Substitute the x-values found in Step 5 into the original equation y = x^3 - 8x^2 - 7x to find the corresponding y-values:
For x = 0:
y = (0)^3 - 8(0)^2 - 7(0) = 0
For x = 4:
y = (4)^3 - 8(4)^2 - 7(4) = 32 - 128 - 28 = -124
So, the two points of intersection are (0, 0) and (4, -124).
Step 7: Write the equations of the tangent lines
The equations of the tangent lines passing through the origin are given by the points of intersection found in Step 6:
1. For the point (0, 0):
The slope of the tangent line is 0/0, which is undefined. In this case, the tangent line is a vertical line passing through the origin. The equation of the line is x = 0.
2. For the point (4, -124):
The slope of the tangent line can be found by substituting the x-value into the derivative equation:
y' = 3x^2 - 16x - 7
y' = 3(4)^2 - 16(4) - 7
y' = 48 - 64 - 7
y' = -23
Since the slope is -23 and the point (4, -124) lies on the line, the equation of the line is:
y = -23x
Therefore, the equations of all lines through the origin that are tangent to the curve y = x^3 - 8x^2 - 7x are: x = 0 and y = -23x.
To find the equations of lines that are tangent to a curve, passing through the origin, we need to find the points on the curve where the slope is equal to zero. Then we can use the equation of a line, which is y = mx, where m is the slope of the line.
Let's find the points (x, y) on the curve where the slope is zero. The slope is zero when the derivative of the curve, dy/dx, is zero. So, we need to find the values of x that satisfy dy/dx = 0.
First, let's find dy/dx by taking the derivative of the curve equation, y = x^3 - 8x^2 - 7x, with respect to x.
dy/dx = 3x^2 - 16x - 7.
Now we set dy/dx equal to zero and solve for x:
3x^2 - 16x - 7 = 0.
We can solve this quadratic equation to find the values of x:
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a),
where a = 3, b = -16, and c = -7, we can calculate x:
x = (-(-16) ± √((-16)^2 - 4(3)(-7))) / (2(3)),
x = (16 ± √(256 + 84)) / 6,
x = (16 ± √340) / 6.
Simplifying further, we have:
x = (16 ± 2√85) / 6,
x = (8 ± √85) / 3.
Now we have the x-values where the slope is zero: x = (8 ± √85) / 3.
To find the corresponding y-values, we substitute these x-values back into the original curve equation:
For x = (8 + √85) / 3, y = ((8 + √85) / 3)^3 - 8((8 + √85) / 3)^2 - 7((8 + √85) / 3).
For x = (8 - √85) / 3, y = ((8 - √85) / 3)^3 - 8((8 - √85) / 3)^2 - 7((8 - √85) / 3).
Therefore, the equations of the lines tangent to the curve, passing through the origin, are:
1) y = (3x - 2√85 - 8) / 3, for x = (8 + √85) / 3.
2) y = (3x + 2√85 - 8) / 3, for x = (8 - √85) / 3.
These are the equations of the lines that are tangent to the curve y = x^3 - 8x^2 - 7x and pass through the origin.