Finding Points with a Given Gradient | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Put a given gradient into the derivative

If the point is given, substitute its x-coordinate into $f'(x)$ . If the gradient is given instead, solve $f'(x)=\text{given gradient}$ to find the possible x-coordinates.

Worked example

Find the points on $y=x^3+3x^2-9x+1$ where the gradient is $15$ .

\begin{gathered}\frac{dy}{dx}=3x^2+6x-9,\\3x^2+6x-9=15,\\x^2+2x-8=0.\end{gathered}

The calculator's polynomial solver gives $2,-4$ . These convenient roots give short exact working:

(x-2)(x+4)=0.

Both values are allowed. Substitute into the original curve, not the derivative, to obtain $(2,3)$ and $(-4,21)$ . Substituting each x-value into the derivative checks that both gradients are $15$ .

For awkward decimal roots, use the quadratic formula to keep an exact answer when required. A calculator root identifies a value; the derivative equation and written solving method still support it.

Extract a gradient from a line

Parallel lines have the same gradient. Rearrange a given line into $y=mx+c$ to read $m$ . For example, $x-2y=5$ becomes $y=\tfrac12x-\tfrac52$ , so a parallel tangent requires $f'(x)=\tfrac12$ .

If the line is given by points $A,B$ , use $m_{AB}=(y_B-y_A)/(x_B-x_A)$ first. A gradient at an x-axis crossing needs $y=0$ to locate the point; a gradient at a y-axis crossing uses $x=0$ .

Find every possible point

Your turnIndependent practice

Find the points on $y=x+4/x$ whose tangents are parallel to $x-2y=5$ . Give exact coordinates.

Show worked answer

The domain excludes $x=0$ . The required gradient is $1/2$ , so

\begin{aligned}1-\frac4{x^2}&=\frac12,\\x^2&=8,\\x&=\pm2\sqrt2.\end{aligned}

Taking the square root gives two real values here; neither is excluded. Substitute into $y=x+4/x$ :

\begin{gathered}(2\sqrt2,3\sqrt2),\\(-2\sqrt2,-3\sqrt2).\end{gathered}

Both give $1-4/8=1/2$ in the derivative.