Find the equation of the line tangent to the given curve with the given slope
y=(1/3)x^3+2x^2+5, m=-4
I'm stuck. So far, I have:
y'=x^2+4x
x^2+4x=-4
y' = x^2 + 4 x = slope of curve = -4 at our spot
so
x^2 + 4 x + 4 = 0
(x+2)^2 = 0
x = -2
so when x = -2
y = -8/3 + 8 + 5 = (-8 +39)/3 = 31/3
so line of slope = -4 through (-2 , 31/3)
y = -4 x + b
31/3 = -4(-2) + b
b = 31/3 - 24/3 = 7/3
y = -4 x + 7/3
3 y = -12 x + 7
the line tangent is y=mx+b
you know m=-4
y=-4x+b is the line.
Now, we know dy/dx=4 at the tangent point, or
4=x^2+4x
(x^2+4x+4)=0 or (x+2)^2=0 or x=-2
and y=(1/3)x^3+2x^2+5= (1/3)(-2)^3+2(-2)^2 +5=10.33.
so the point of tangency is -2,10.33
y=mx+b
10.33=-4(-2)+b solve for b, and you have the equation of the line.
To find the equation of the line tangent to a curve with a given slope, you can follow these steps:
1. Start with the equation of the curve given. In this case, the curve equation is y = (1/3)x^3 + 2x^2 + 5.
2. Take the derivative of the curve equation to find the slope of the tangent line. The derivative of the equation y = (1/3)x^3 + 2x^2 + 5 is y' = x^2 + 4x.
3. Set the derivative equal to the given slope, in this case, -4:
x^2 + 4x = -4
4. Rearrange the equation to solve for x. Adding 4 to both sides and simplifying the equation:
x^2 + 4x + 4 = 0
(x + 2)^2 = 0
5. Solve for x by taking the square root of both sides:
x + 2 = 0
x = -2
6. Plug the value of x into the original curve equation to find the corresponding y-value:
y = (1/3)(-2)^3 + 2(-2)^2 + 5
y = -2/3 + 8 + 5
y = 21/3
y = 7
7. Now you have the point of tangency, (-2, 7), and the slope of the line, -4. Use the point-slope form of a line to find the equation:
y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope.
y - 7 = -4(x + 2)
y - 7 = -4x - 8
y = -4x - 1
Therefore, the equation of the line tangent to the curve y = (1/3)x^3 + 2x^2 + 5 with a slope of -4 is y = -4x - 1.