Consider the parabola y = 6x − x2.
(a) Find the slope of the tangent line to the parabola at the point (1, 5).
4
(b) Find an equation of the tangent line in part (a).
y =
(a) To find the slope of the tangent line at the point (1, 5), we need to find the derivative of the parabola.
The equation of the parabola is y = 6x - x^2.
To find the derivative, we differentiate the equation with respect to x:
dy/dx = d(6x - x^2)/dx = 6 - 2x.
Now, substitute x = 1 into the derivative:
dy/dx = 6 - 2(1) = 6 - 2 = 4.
Therefore, the slope of the tangent line at the point (1, 5) is 4.
(b) Now, we need to find an equation of the tangent line.
We have the slope of the tangent line, which is 4, and the point (1, 5) that the line passes through.
Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we substitute the values:
y - 5 = 4(x - 1).
Expanding and simplifying the equation, we get:
y - 5 = 4x - 4.
Add 5 to both sides to isolate y:
y = 4x + 1.
Therefore, the equation of the tangent line at the point (1, 5) is y = 4x + 1.
To find the slope of the tangent line to the parabola at the point (1, 5), we need to find the derivative of the equation of the parabola.
(a) Finding the derivative:
To find the derivative of the parabola y = 6x − x^2, we can use the power rule. The power rule states that the derivative of x^n, where n is a constant, is equal to n * x^(n-1).
For the equation y = 6x − x^2, we can find the derivative as follows:
dy/dx = d/dx(6x − x^2)
= d/dx(6x) - d/dx(x^2)
= 6 - 2x
(b) Finding the slope:
Now that we have the derivative, we can substitute x = 1 into the derivative expression to find the slope of the tangent line at the point (1, 5).
dy/dx = 6 - 2x
= 6 - 2(1)
= 6 - 2
= 4
So, the slope of the tangent line to the parabola at the point (1, 5) is 4.
Now, to find an equation of the tangent line, we can use the point-slope form of a linear equation, which is y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope of the tangent line.
Using the point (1, 5) and the slope 4, we substitute these values into the point-slope form:
y - 5 = 4(x - 1)
Simplifying the equation, we get:
y - 5 = 4x - 4
Rearranging the equation to solve for y, we have:
y = 4x + 1
So, the equation of the tangent line at the point (1, 5) is y = 4x + 1.