dy/dx = a
y=ax y'=ln(a)*ax
If y = ax and a = 0, then y=0. No matter what x is, y is still 0. Therefore any graph of y=ax (where a=0) will simply be a line at y=0, which is the x-axis.
Ax + By = C By = -Ax + C y = (-A/B)x + C/B
a function
Ax + By = CSubtract 'Ax' from each side:By = -Ax + CDivide each side by 'B' :y = -(A/B)x + C/BThe slope of the line is -(A/B) .The y-intercept is (C/B) .
y=ax y'=ln(a)*ax
If you mean: y = x - a/x Then: y = x - ax-1 y' = 1 + ax-2 y' = 1 + a/x2 If you mean: y = (x - a)/x Then: y = 1 - ax-1 y' = ax-2 y' = a/x2
If y = ax and a = 0, then y=0. No matter what x is, y is still 0. Therefore any graph of y=ax (where a=0) will simply be a line at y=0, which is the x-axis.
Ax + By = C By = -Ax + C y = (-A/B)x + C/B
a function
14
y' = (sec(x))^2
Ax + By = CSubtract 'Ax' from each side:By = -Ax + CDivide each side by 'B' :y = -(A/B)x + C/BThe slope of the line is -(A/B) .The y-intercept is (C/B) .
To solve for B in the equation ( Ax + By = C ), you first isolate the term involving B. Rearranging gives ( By = C - Ax ). Then, divide both sides by y (assuming y is not zero) to solve for B: ( B = \frac{C - Ax}{y} ).
1
13
The graph of the first form passes through the origin while the second does not - unless c = 0.