Find the slope of the tangent to the curve y=2sqrt(x) at the point where x=a.
Find equations of the tangent lines at the points (1, 2) and (9, 6).
y = 2√x
y' = 1/√x
so at the point (a,2√a) the tangent line has equation
y-2√a = 1/√a (x-a)
so plug in whatever numbers you like
To find the slope of the tangent to the curve y=2√(x) at the point where x=a, you'll need to use the derivative of the function.
Step 1: Find the derivative of y with respect to x.
Differentiate y=2√(x) with respect to x using the power rule and chain rule:
dy/dx = (2/2) * (1/2) * x^(-1/2) = x^(-1/2)
Step 2: Substitute the x-coordinate at which you want to find the slope.
To find the slope of the tangent at point (a, y), substitute x=a into the derivative equation:
dy/dx = a^(-1/2)
Therefore, the slope of the tangent to the curve y=2√(x) at the point where x=a is a^(-1/2).
Now, let's find the equations of the tangent lines at the points (1, 2) and (9, 6).
To find the equation of a tangent line, we need the slope of the tangent and a point on the curve.
For the tangent line at point (1, 2):
Step 1: Find the slope of the tangent.
Substitute x=1 into the derivative equation:
dy/dx = 1^(-1/2) = 1
Step 2: Use the point-slope form of a line.
Using the point-slope form y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line:
y - 2 = 1(x - 1)
y - 2 = x - 1
y = x + 1
Therefore, the equation of the tangent line at the point (1, 2) is y = x + 1.
For the tangent line at point (9, 6):
Step 1: Find the slope of the tangent.
Substitute x=9 into the derivative equation:
dy/dx = 9^(-1/2) = 1/3
Step 2: Use the point-slope form of a line.
Using the point-slope form y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line:
y - 6 = (1/3)(x - 9)
y - 6 = (1/3)x - 3
y = (1/3)x + 3
Therefore, the equation of the tangent line at the point (9, 6) is y = (1/3)x + 3.