a. find an equation for the secant line through the points where x has the given values.
b. find a equation for the line tangent to the curve when x has the first value.
y=9square root(x); x=16, x=25
y = 9√x
y(16) = 36
y(25) = 45
So, the secant has slope (45-36)/(25-16) = 1
So, now use the point-slope form to get the equation for a line with slope 1 through (16,36)
For the tangent line, the slope
y' = 4.5/√x is 9/8 at x=16
Again, use the point-slope form to get the tangent line
a. To find the equation for the secant line through the points where x has the given values, we can use the slope-intercept form of a linear equation, which is y = mx + b.
First, we need to find the coordinates of the two points on the curve:
For x = 16:
y = 9√(16) = 9 * 4 = 36
So, the first point is (16, 36).
For x = 25:
y = 9√(25) = 9 * 5 = 45
So, the second point is (25, 45).
Next, let's calculate the slope (m) using the formula:
m = (y2 - y1) / (x2 - x1)
Using the coordinates of the two points:
m = (45 - 36) / (25 - 16)
m = 9 / 9
m = 1
Now that we have the slope, we can substitute one of the points (let's use the first point) and the slope into the slope-intercept form:
y = mx + b
36 = 1 * 16 + b
36 = 16 + b
b = 36 - 16
b = 20
Therefore, the equation for the secant line through the points where x = 16 and x = 25 is y = x + 20.
b. To find the equation for the line tangent to the curve when x has the first value (x = 16), we need to find the derivative of the curve at that point.
Given: y = 9√(x)
To find the derivative, we can use the power rule for differentiation:
d/dx [ 9√(x) ] = 9 * (1/2) * x^(-1/2) = 4.5√(x^(-1))
Now let's evaluate the derivative at x = 16:
d/dx [ 9√(x) ] = 4.5√(16^(-1)) = 4.5 * (1/4) = 1.125
So, the slope of the tangent line at x = 16 is 1.125.
Now, let's use the point-slope form of a linear equation to find the equation for the tangent line:
y - y1 = m(x - x1)
Using the coordinates (16, 36) and the slope (m = 1.125):
y - 36 = 1.125(x - 16)
Expanding and simplifying:
y - 36 = 1.125x - 18
y = 1.125x - 18 + 36
y = 1.125x + 18
Therefore, the equation for the line tangent to the curve when x = 16 is y = 1.125x + 18.
a. To find the equation for the secant line through the points where x has the given values, we need to first find the corresponding y-values for these x-values.
Given the function y = 9√(x), we can substitute the given x-values to find the corresponding y-values:
For x = 16:
y = 9√(16) = 9 * 4 = 36
For x = 25:
y = 9√(25) = 9 * 5 = 45
Now we have two points on the secant line: (16, 36) and (25, 45).
The equation for a secant line passing through two points (x1, y1) and (x2, y2) can be found using the slope-intercept form equation:
y - y1 = [(y2 - y1) / (x2 - x1)] * (x - x1)
Let's substitute the values into the equation:
For (16, 36):
y - 36 = [(45 - 36) / (25 - 16)] * (x - 16)
Simplifying it further:
y - 36 = (9 / 9) * (x - 16)
y - 36 = x - 16
Final equation for the secant line passing through (16, 36) and (25, 45):
y = x + 20
b. To find the equation for the line tangent to the curve when x has the first value, we need to find the derivative of the given function and evaluate it at the given x-value.
Given the function y = 9√(x), let's find its derivative:
dy/dx = (1/2) * 9 * x^(-1/2)
= 4.5 / √(x)
Now, substitute the given x-value of x = 16 into the derivative:
dy/dx = 4.5 / √(16)
= 4.5 / 4
= 1.125
The slope of the tangent line at x = 16 is 1.125.
Using the point-slope form equation of a line, we can write the equation of the tangent line:
y - y1 = m * (x - x1)
Substituting the values into the equation, with (x1, y1) = (16, 36) and m = 1.125:
y - 36 = 1.125 * (x - 16)
Simplifying it further:
y - 36 = 1.125x - 18
y = 1.125x + 18
Final equation for the tangent line when x = 16:
y = 1.125x + 18