Consider the parabola y = 7x - x2.
(a) Find the slope of the tangent line to the parabola at the point (1, 6).
1
1 is not correct.
Consider f(x)=7x-x²
f'(x)=7-2x
at the point (1,6), x=1, so the slope required is f'(1).
To find the slope of the tangent line to the parabola at the point (1, 6), we need to take the derivative of the equation of the parabola with respect to x.
The given equation of the parabola is y = 7x - x^2.
Taking the derivative of y with respect to x:
dy/dx = d/dx (7x - x^2)
= 7 - 2x
Now, plug in the x-coordinate of the given point, which is x = 1, into the derived equation:
dy/dx = 7 - 2(1)
= 7 - 2
= 5
Therefore, the slope of the tangent line to the parabola at the point (1, 6) is 5.
To find the slope of the tangent line to the parabola at the point (1, 6), we need to find the derivative of the equation of the parabola and evaluate it at x = 1.
Given the equation of the parabola, y = 7x - x^2, we can find the derivative by applying the power rule for derivatives.
The power rule states that if you have a function f(x) = ax^n, then its derivative is given by f'(x) = nax^(n-1).
Applying the power rule to our equation y = 7x - x^2, we have:
dy/dx = d/dx (7x - x^2)
= 7*1 - 2x^(2-1)
= 7 - 2x
Now, to find the slope at x = 1, we substitute x = 1 into the derivative equation:
dy/dx = 7 - 2(1)
= 7 - 2
= 5
Therefore, the slope of the tangent line to the parabola at the point (1, 6) is 5.