To find the gradient of a function, you calculate the partial derivatives of the function with respect to each variable. For a function ( f(x, y) ), the gradient is represented as a vector ( \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) ). This vector points in the direction of the steepest ascent of the function and its magnitude indicates the rate of increase. You can compute the gradient using calculus techniques, such as differentiation.
(-1.5,0) (1.5,0) what is the gradient?
Draw a tangent to the curve at the point where you need the gradient and find the gradient of the line by using gradient = up divided by across
i think you do Vertical/horizontal
divide by the gradient
y=mx+c m=gradient c= is the y intercept in ur case: y=5x so gradient = 5
find the gradient
(-1.5,0) (1.5,0) what is the gradient?
Draw a tangent to the curve at the point where you need the gradient and find the gradient of the line by using gradient = up divided by across
i think you do Vertical/horizontal
divide by the gradient
y=mx+c m=gradient c= is the y intercept in ur case: y=5x so gradient = 5
To find the gradient on a quadratic graph, you first need to determine the derivative of the quadratic function, which is typically in the form (y = ax^2 + bx + c). The derivative, (y' = 2ax + b), represents the gradient at any point (x) on the curve. By substituting a specific (x) value into the derivative, you can find the gradient at that particular point on the graph. This gradient indicates the slope of the tangent line to the curve at the chosen point.
Yes beccause: (y1-y2)/(x1-x2) = gradient
english?
If A = (xa, ya) and B = (xb, yb) and xa is not equal to xb, then gradient of AB = (ya - yb)/(xb - xb).If xa = xb then the gradient is undefined.
If 6x-2y=18 then -2y=-6x+18 and y=3x+9 so the gradient is 3
Select two points on the graph and suppose their coordinates are (x1, y1) and (x2, y2) then the gradient = (y1 - y2) / (x1 - x2) provided that x1 and x2 are different. If not, the gradient is not defined.