Any and all help would be appreciated.
Find the equation of the tangent line to the curve y = (x - 2)^5 (x + 1)^2 at the point (3, 16)
The line passing through a point Pt(xt,yt) is
(y-yt) = m(x-xt)
where m is the slope.
For the given function
f(x) = (x - 2)^5 (x + 1)^2
We first validate that f(3) = 16, the tangent point.
f(3) = (3 - 2)^5 (3 + 1)^2 =16 ..... OK
The only unknown is the slope m, which we should be able to find by the derivative of the function, f'(x).
Differentiating f(x) as the product of two binomials, (x-2)5 and (x+1)2, and utilising the product rule, d(uv)=udv+vdu, we get
f'(x) = (x-2)5*2(x+1) + 5(x-2)4(x+1)2
The slope m can be found readily by
m=y'(3)
=88
Thus the equation of the required tangent is:
(y-yt) = m(x-xt)
(y-16) = 88(x-3)
y=88x-248
Take the derivative dy/dx (that is the slope).
y'=5(x-2)^4(x+1)^2+2(x-2)^5((x+1)
Put in the point x=3
Now you have the slope at x=3
y=mx+b
you have y, m, x...solve for b.
To find the equation of the tangent line to the curve, we can use the derivative of the function at the given point. The derivative gives us the slope of the tangent line at any point on the curve.
First, we need to find the derivative of the function y = (x - 2)^5 (x + 1)^2. To do this, we'll apply the product rule and then algebraically simplify the expression.
Let's start by finding the derivative of the first term, (x - 2)^5. Using the power rule, the derivative of (x - 2)^5 is 5(x - 2)^4.
Next, let's find the derivative of the second term, (x + 1)^2. Again using the power rule, the derivative of (x + 1)^2 is 2(x + 1)^1.
Applying the product rule, the derivative of the whole function is:
dy/dx = [(x - 2)^5] * 2(x + 1) + [(x + 1)^2] * 5(x - 2)^4
Simplifying this expression, we get:
dy/dx = 2(x - 2)^5(x + 1) + 5(x + 1)^2(x - 2)^4
Evaluate the derivative at the given point (3, 16) by substituting x = 3 into the expression.
dy/dx = 2(3 - 2)^5(3 + 1) + 5(3 + 1)^2(3 - 2)^4
= 2(1)^5(4) + 5(4)^2(1)^4
= 2(1)(4) + 5(16)(1)
= 8 + 80
= 88
So, the derivative of the function at the point (3, 16) is 88, which is the slope of the tangent line at that point.
Now that we have the slope of the tangent line, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line and m is the slope of the line.
Substituting the values x1 = 3, y1 = 16, and m = 88 into the equation, we get:
y - 16 = 88(x - 3)
Simplifying this equation, we obtain:
y - 16 = 88x - 264
Rearranging the equation to the standard form, we have:
y = 88x - 248
Therefore, the equation of the tangent line to the curve y = (x - 2)^5 (x + 1)^2 at the point (3, 16) is y = 88x - 248.