Find the point(s) on the curve \(y=x^3\) where the line through the point \((4,0 ... Consider the function \(\ds y=3x^2+2x-10\) on the interval \([1,5]\text{.}\) Does this function have a tangent line ...

We can calculate the gradient of a tangent to a curve by differentiating ... curve to find the y-coordinate Substitute your point on the line and the gradient into \(y - b = m(x - a)\) Find ...

As a tangent is a straight line it is described by an equation in the form \(y - b = m(x - a)\). You need both a point and the gradient to find its equation. You are usually given the point - it's ...