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.
y = (c - Ax) / (B)
a function
y' = (sec(x))^2
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.
y = (c - Ax) / (B)
a function
y' = (sec(x))^2
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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) .
1
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The graph of the first form passes through the origin while the second does not - unless c = 0.
Solve the equation for y. This will give you an equation similar to y = ax + b, where a is the slope, and b is the y-intercept.