Extension (Sec 3/4): this section goes beyond the Sec 2 syllabus. A graph of y=f(x) can solve other equations: the roots of f(x)=k are the x-coordinates where the curve meets the line y=k, and the roots of f(x)=mx+c are where it meets the straight line y=mx+c. Why: at an intersection both equations are true at once. On y=x2-x-2, the line y=x+1 meets the curve where x2-x-2=x+1, that is x2-2x-3=0, at x=-1 and x=3. The dashed line y=2 gives the roots of x2-x-4=0, about -1.6 and 2.6. Method - Rearrange the new equation so that one side is exactly the expression already drawn, for example x2-x-2=x+1. - Draw the line given by the other side, using two or three points. - Read the x-coordinates of the intersections to the accuracy the scale allows. - Gradient of the curve at a point: draw the tangent there with a ruler, pick two far-apart points on it and work out rise run. Example Use the graph of y=x2-x-2 to solve x2-2x-3=0. x2-2x-3 = 0 x2-x-2 = x+1 Draw y=x+1 through (-1,0) and (3,4). The intersections give x=-1 or x=3. Gradient: on y=x2-x-2, a tangent drawn at (2,0) passes through about (1,-3) and (3,3), so the gradient there is about 3-(-3)3-1=3. Check: (-1)2-2(-1)-3=0 and 9-6-3=0. A common error is to draw y=-2x-3 instead; the line must make the two sides of the rearranged equation equal.

Solving equations from graphs and finding gradients (extension)

Secondary 2

Extension (Sec 3/4): this section goes beyond the Sec 2 syllabus. A graph of $y=f(x)$ can solve other equations: the roots of $f(x)=k$ are the $x$-coordinates where the curve meets the line $y=k$, and the roots of $f(x)=mx+c$ are where it meets the straight line $y=mx+c$.

Why: at an intersection both equations are true at once. On $y=x^2-x-2$, the line $y=x+1$ meets the curve where $x^2-x-2=x+1$, that is $x^2-2x-3=0$, at $x=-1$ and $x=3$. The dashed line $y=2$ gives the roots of $x^2-x-4=0$, about $-1.6$ and $2.6$.

Method

Rearrange the new equation so that one side is exactly the expression already drawn, for example $x^2-x-2=x+1$.
Draw the line given by the other side, using two or three points.
Read the $x$-coordinates of the intersections to the accuracy the scale allows.
Gradient of the curve at a point: draw the tangent there with a ruler, pick two far-apart points on it and work out rise $\div$ run.

Example

Use the graph of $y=x^2-x-2$ to solve $x^2-2x-3=0$.

$\begin{aligned} x^2-2x-3 &= 0 \\ x^2-x-2 &= x+1 \end{aligned}$

Draw $y=x+1$ through $(-1,0)$ and $(3,4)$. The intersections give $x=-1$ or $x=3$.

Gradient: on $y=x^2-x-2$, a tangent drawn at $(2,0)$ passes through about $(1,-3)$ and $(3,3)$, so the gradient there is about $\dfrac{3-(-3)}{3-1}=3$.

Check: $(-1)^2-2(-1)-3=0$ and $9-6-3=0$. A common error is to draw $y=-2x-3$ instead; the line must make the two sides of the rearranged equation equal.

Create a free account to practise similar questions.

Create free account Sign in Get the E Math course · from S$29.98/mo E Math tuition

Need help? Join our E-Maths tuition for O-Level success.

Learn more