Using graphs to solve equationsAQA A-Level Maths: Revision notes

A solution of $f(x)=0$ is a value of $x$ where the graph of $y=f(x)$ meets the $x$-axis, because $y=0$ there. The number of real solutions equals the number of places the curve meets the axis. For $y=x^2-4x-5=(x-5)(x+1)$ the curve meets the axis at $x=-1$ and $x=5$, so $x^2-4x-5=0$ has these two solutions. For $y=x^3-3x$, the curve meets the axis at $x=0,\pm\sqrt3$, three solutions. A graph that touches the axis shows a repeated root; one that never reaches the axis shows no real solutions.

Where the graphs of $y=f(x)$ and $y=g(x)$ meet, both have the same $x$ and $y$. So the $x$-coordinates of the intersections are the solutions of $f(x)=g(x)$, and the $y$-coordinates are found by substituting back. Example: the curve $y=x^2-4x-5$ and the line $y=2x-5$. Equating: $x^2-4x-5=2x-5$, so $x^2-6x=0$ and $x=0$ or $x=6$. Substituting into the line: $(0,-5)$ and $(6,7)$. This is the same as solving the simultaneous equations $y=f(x)$ and $y=g(x)$.

To solve an equation using a graph that is already drawn, rearrange it so one side is the drawn function. What is left on the other side is the line to draw.

• Given $y=x^2-4x-5$, to solve $x^2-4x-3=0$ write $x^2-4x-5=-2$, so draw $y=-2$.
• Given $y=x^2-2x-8$, to solve $x^2-3x-10=0$ write $x^2-2x-8=x+2$, so draw $y=x+2$. The solutions are the $x$-coordinates of the points where the new line meets the curve. Check by solving the original equation algebraically.

For a curve $y=f(x)$, the number of solutions of $f(x)=k$ is the number of times the horizontal line $y=k$ meets the curve. For $y=x^3-3x$, with maximum $(-1,2)$ and minimum $(1,-2)$:

• $|k|>2$: one solution;
• $k=2$ or $k=-2$: two distinct solutions (one repeated);
• $|k|<2$: three solutions. The turning points set where the number of solutions changes.

Equate a line and a curve to obtain a quadratic and look at its discriminant $b^2-4ac$:

• $b^2-4ac>0$: two intersection points (the line cuts the curve);
• $b^2-4ac=0$: one repeated root, so the line is a tangent;
• $b^2-4ac<0$: no real roots, so the line does not meet the curve. Example: $y=x^2-5x+2$ and $y=x-7$ give $x^2-6x+9=(x-3)^2=0$, so the line is a tangent at $x=3$. For $y=x+c$ the equation is $x^2-6x+(2-c)=0$ and $36-4(2-c)<0$ gives $c<-7$ for no intersection.