find the slope of the tangent line
y= -5x^1/2 + X^3/2; X=25
dy/dx = (-5/2)x^(-1/2) + (3/2)x^(1/2)
= -5/(2√x) + 3√x/2
when x = 25
y = -5(√25) + (√25)^3 = -25 + 125 = 100
dy/dx = -5/(2√25) + 3√25/2 = -1/2 + 15/2 = 7
then
(y-100) = 7(x-25)
Finish it up to express the final equation is the form you want.
To find the slope of the tangent line at a given point on a curve, you need to take the derivative of the function and then substitute the x-coordinate of the point into the derivative.
Let's find the derivative of the given function, y = -5x^(1/2) + x^(3/2):
First, we differentiate each term separately.
For the first term, -5x^(1/2), we use the Power Rule for differentiation. The Power Rule states that when differentiating a function of the form f(x) = ax^n, the derivative is given by f'(x) = anx^(n-1).
So, for -5x^(1/2), the derivative is (-5)(1/2)x^((1/2)-1) = -2.5x^(-1/2).
For the second term, x^(3/2), we apply the Power Rule again. The derivative is (3/2)x^((3/2)-1) = (3/2)x^(1/2).
Now, we can find the derivative of the entire equation by adding the derivatives of the individual terms:
dy/dx = -2.5x^(-1/2) + (3/2)x^(1/2)
To find the slope of the tangent line at a specific point, substitute the x-coordinate of that point into the derivative. In this case, the given x-coordinate is X = 25:
dy/dx = -2.5(25)^(-1/2) + (3/2)(25)^(1/2)
Now, simplify the equation by evaluating the exponentials:
dy/dx = -2.5/√25 + 3/2√25
dy/dx = -2.5/5 + 3/10
dy/dx = -0.5 + 0.3
dy/dx = -0.2
Therefore, the slope of the tangent line at the point where x = 25 is -0.2.