--- Find the slope m of the tangent to the curve y = 4 + 4x^2 − 2x^3 at the point where x = a.
---- Find equations of the tangent lines at the points (1,6) and (2,4).
(1,6) Y(x)=
(2,4) Y(x)=
y' = 8x-6x^2
so, at x=a, y' = 8a-6a^2 = m
Use the above info to plug into the point-slope form of the lines. No sweat.
To find the slope of the tangent to the curve at a given point, you need to find the derivative of the curve and evaluate it at that particular point. Once you have the slope, you can use the point-slope form of a line to find the equation of the tangent line.
Let's go through the steps to find the slope m of the tangent to the curve y = 4 + 4x^2 - 2x^3 at the point where x = a:
Step 1: Find the derivative of the curve.
To find the derivative, we need to apply the power rule for derivatives. For each term, multiply the coefficient by the exponent of x, then reduce the exponent by 1. For the given curve, the derivative will be:
y' = d/dx (4 + 4x^2 - 2x^3)
= 0 + (2 * 4 * x^(2-1)) - (3 * 2 * x^(3-1))
= 8x - 6x^2
Step 2: Evaluate the derivative at the point x = a.
Replace every occurrence of x in the derivative with the value a:
m = 8a - 6a^2
So, the slope of the tangent to the curve at the point (x = a) is given by 8a - 6a^2.
Now let's find the equations of the tangent lines at the points (1,6) and (2,4):
To find the equation of a line, we need the slope (m) and the coordinates of a point (x, y) on the line (x1, y1). We can use the point-slope form of a line:
y - y1 = m(x - x1)
For the point (1,6):
m = slope at x = 1 = 8(1) - 6(1^2) = 8 - 6 = 2
(x1, y1) = (1, 6)
Using the point-slope form:
y - 6 = 2(x - 1)
y - 6 = 2x - 2
y = 2x + 4
So, the equation of the tangent line at the point (1,6) is y = 2x + 4.
For the point (2,4):
m = slope at x = 2 = 8(2) - 6(2^2) = 16 - 24 = -8
(x1, y1) = (2, 4)
Using the point-slope form:
y - 4 = -8(x - 2)
y - 4 = -8x + 16
y = -8x + 20
So, the equation of the tangent line at the point (2,4) is y = -8x + 20.