Tangents and normalsAQA A-Level Maths: Revision notes

The gradient of a curve at a point is the gradient of the tangent there, found by evaluating the derivative at that point. For $y=x^3-2x$, $\frac{dy}{dx}=3x^2-2$, so at $x=2$ the gradient is $3(4)-2=10$. Always differentiate first, then substitute the $x$-coordinate; substituting into the equation of the curve gives the value of $y$, not the gradient.

A tangent is a straight line, so use $y-y_1=m(x-x_1)$ with $m=\frac{dy}{dx}$ at $x_1$ and $(x_1,y_1)$ a point on the curve. Example: $y=2x^3-5x+1$ at $x=2$. Then $y_1=16-10+1=7$ and $m=6(4)-5=19$, so $y-7=19(x-2)$, i.e. $y=19x-31$. Find $y_1$ from the curve, not from the gradient.

The normal at a point is the straight line through that point perpendicular to the tangent. Perpendicular gradients satisfy $m_1m_2=-1$, so if the tangent has gradient $m$ the normal has gradient $-\frac1m$. For $y=2x^3-5x+1$ at $(2,7)$ the normal gradient is $-\frac1{19}$: $y-7=-\frac1{19}(x-2)$, which rearranges to $x+19y-135=0$. If the tangent is horizontal the normal is vertical ($x=k$), and if the tangent is vertical the normal is horizontal.

To find where a tangent is parallel to a given line, set $\frac{dy}{dx}$ equal to that line's gradient and solve for $x$, then find $y$ from the curve. For $y=x^3-3x^2-9x+4$ and a line $y=15x-2$: $3x^2-6x-9=15$, so $x^2-2x-8=0$, giving $x=4$ or $x=-2$ and the points $(4,-16)$ and $(-2,2)$. A tangent parallel to the $x$-axis has gradient $0$; one perpendicular to a line of gradient $m$ has gradient $-\frac1m$.

To find where a tangent or normal meets an axis, set $y=0$ (for the $x$-axis) or $x=0$ (for the $y$-axis) in its equation. To find where it meets the curve again, equate the line to the curve's equation. The point of contact is already one root of the resulting equation, so factorise it out; for a tangent it is a repeated root. Example: the normal to $y=x^2+3x-2$ at $(1,2)$ is $y=-\frac15x+\frac{11}5$. Equating gives $5x^2+16x-21=0$, so $(5x+21)(x-1)=0$ and the other point has $x=-\frac{21}5$.

When a curve contains unknown constants, use two conditions: the point lies on the curve (substitute its coordinates) and the gradient there is known (substitute into $\frac{dy}{dx}$). For $y=x^2+px+q$ with gradient $5$ at $(1,2)$: $2+p=5$ gives $p=3$, and $1+3+q=2$ gives $q=-2$.