Qn: the equation of the tangent to the curve y=kx+(6/x) at the point (-2,-19) is px + qy = c
find the values of k, p, q and c
Bailey, Andrea, Danny, Riley, anabelle, holly, LILLY, zachary, I'm stumped -- or whoever!
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step 1
sub the point (-2,-19) into
y = kx - 6/x and solve for k (I got k=8)
step 2. Now that you know the actual function, find its derivative,
I got dy/dx = 8 - 6/x^2
sub in x=-2 to get the slope of the tangent, I got slope = 13/2
Step3. Now that you have the slope of the line and the point on the line,
find the equation.
I got 13x - 2y = c
sub in the point
-26 + 38 = c = 12
compare
13x - 2y = 12 with
px + qy = c
To find the equation of the tangent to the curve y = kx + (6/x) at the point (-2, -19), we need to use the concept of differentiation.
Step 1: Differentiate the equation y = kx + (6/x) with respect to x to find the derivative of the curve.
The derivative of kx is k and the derivative of 6/x is -6/x^2 (using the power rule and chain rule).
Therefore, the derivative of y = kx + (6/x) is dy/dx = k - 6/x^2.
Step 2: Substitute the x-coordinate of the given point (-2, -19) into dy/dx to find the slope of the tangent line at that point.
dy/dx = k - 6/(-2)^2 = k - 6/4 = k - 3/2
The slope of the tangent line at (-2, -19) is given by k - 3/2.
Step 3: The equation of a line in point-slope form is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Using the point-slope form, we can substitute (-2, -19) as (x1, y1) and k - 3/2 as the slope in the equation.
y - (-19) = (k - 3/2)(x - (-2))
Simplifying the equation gives y + 19 = (k - 3/2)(x + 2).
Step 4: Rearrange the equation to get it into the form px + qy = c.
Expanding the expression on the right side of the equation gives (k - 3/2)(x + 2) = kx - (3/2)x + 2k - 3.
Rearranging the equation further gives kx - (3/2)x + 2k - 3 - y - 19 = 0.
Combining like terms gives kx - (3/2)x - y + 2k - 22 = 0.
Comparing this equation with px + qy = c, we can find the values of p, q, and c.
p = k - 3/2
q = -1
c = 2k - 22
Therefore, the values of k, p, q, and c, respectively, are k, k - 3/2, -1, and 2k - 22.
To find the equation of the tangent to the curve at a given point, we need to find the slope of the curve at that point. The slope of a curve at a given point can be found by taking the derivative of the equation of the curve with respect to x.
Let's start by finding the derivative of the equation y = kx + (6/x).
dy/dx = d/dx (kx + 6/x)
Applying the power rule of differentiation, we get:
dy/dx = k - 6/x^2
Since the derivative gives us the slope of the curve at any given point, we can substitute the x-coordinate of the given point (-2) into the derivative to find the slope at that point:
dy/dx = k - 6/(-2)^2 = k - 6/4 = k - 3/2
Now we know the slope of the tangent at the point (-2, -19). Let's call this slope m:
m = k - 3/2
We also know that the equation of a line can be written in the form of px + qy = c, where p, q, and c are constants.
The equation of the tangent to the curve at (-2, -19) can be written as:
y - (-19) = m(x - (-2))
Simplifying, we get:
y + 19 = m(x + 2)
y + 19 = mx + 2m
mx - y + 2m - 19 = 0
Comparing this equation with the standard form px + qy = c, we can conclude that:
p = m
q = -1
c = 2m - 19
Since we already found that m = k - 3/2, we can substitute this back into the values of p, q, and c to get the final values:
p = k - 3/2
q = -1
c = 2(k - 3/2) - 19