Sketching Quadratics (SQA National 5 Maths): Revision Note

Sketching quadratics y = (ax - m)(bx - n)

What is the connection between the solutions of a quadratic equation and the corresponding quadratic graph?

• There is an important connection between

  • the quadratic graph of $y=a{x}^2+bx+c$

  • and the solutions of the corresponding quadratic equation $a{x}^2+bx+c=0$

• If $x=p$ is a solution to $a{x}^2+bx+c=0$

  • then the graph of $y=a{x}^2+bx+c$ intercepts the $\mathit{y}$-axis at $\left(p,0\right)$

  • Remember that $y=0$ is the equation of the $x$-axis

• This means you can use a quadratic graph to find solutions to the corresponding quadratic equation

  • The $x$-coordinates of its $x$-axis intercepts (if any) are the solutions to the equation

• It also means you can use the solutions to a quadratic equation to determine features of the corresponding quadratic graph

  • For example if the equation has two distinct solutions, then the graph intersects the $x$-axis at two points

  • If the equation only has one solution, then the graph only touches the $x$-axis at a single point

  • If the equation has no solutions, then the graph does not intersect the $x$-axis (it is either wholly above or wholly below it)

How do I sketch a quadratic graph from an equation in factorised form?

• It is easy to sketch the graph of a quadratic if its equation is given in factorised form $y=\left(ax-m\right)\left(bx-n\right)$

• For example, to sketch the graph of $y=\left(2x-1\right)\left(2x-7\right)$

  • Solve $\left(2x-1\right)\left(2x-5\right)=0$ by setting the brackets equal to zero and solving

    • $2x-1=0\Rightarrow x=0.5$

    • $2x-7=0\Rightarrow x=3.5$

    • So the quadratic intercepts the $\mathit{x}$-axis at $\left(0.5,0\right)$ and $\left(3.5,0\right)$

  • The axis of symmetry occurs midway between the $x$-axis intercepts

    • $\frac{0.5+3.5}{2}=\frac{4}{2}=2$

    • So the axis of symmetry is $x=2$

  • The turning point lies on the axis of symmetry

    • Substitute $x=2$ into the equation

    • $y=\left(2\left(2\right)-1\right)\left(2\left(2\right)-7\right)=3\times \left(-3\right)=-9$

    • So the turning point is at $\left(2,-9\right)$

  • The $\mathit{y}$-axis intercept occurs when $x=0$

    • Substitute $x=0$ into the equation

    • $y=\left(2\left(0\right)-1\right)\left(2\left(0\right)-7\right)=\left(-1\right)\times \left(-7\right)=7$

    • So the $y$-axis intercept is at $\left(0,7\right)$

  • That is all you need to sketch the graph!

    • If the turning point is below the $x$-axis (i.e. has a negative $y$-coordinate)

      • then the parabola is $\cup$-shaped

    • If the turning point is above the $x$-axis (i.e. has a positive $y$-coordinate)

      • then the parabola is $\cap$-shaped

Sketching quadratics y = k(x + p)² + q

How do I sketch a quadratic graph from an equation in completed square form?

• It is easy to sketch the graph of a quadratic if its equation is given in completed square form $y=k{\left(x+p\right)}^2+q$

• For example, to sketch the graph of $y=2{\left(x-4\right)}^2+1$

  • The turning point of $y=k{\left(x+p\right)}^2+q$ is at $\left(-p,q\right)$

    • So the turning point of $y=2{\left(x-4\right)}^2+1$ is at $\left(4,1\right)$

  • The $\mathit{y}$-axis intercept occurs when $x=0$

    • Substitute $x=0$ into the equation

    • $y=2{\left(0-4\right)}^2+1=2{\left(-4\right)}^2+1=2\times 16+1=33$

    • So the $y$-axis intercept is at $\left(0,33\right)$

  • The number in front of the bracket, $k$, tells you the shape of the parabola

    • If $k$ is positive, the parabola is $\cup$-shaped

    • If $k$ is negative, the parabola is $\cap$-shaped

    • So $y=2{\left(x-4\right)}^2+1$ is $\cup$-shaped

  • That is all you need to sketch the graph!

    • Note that $y=2{\left(x-4\right)}^2+1$

      • is $\cup$-shaped

      • and has its turning point above the $x$-axis at $\left(0,33\right)$

    • That means that it will not intercept the $x$-axis

• If you need to find $\mathit{x}$-axis intercepts for a quadratic in $y=k{\left(x+p\right)}^2+q$ form, you will need to set the quadratic equal to zero and then solve for $x$

  • For example, for $y=2{\left(x-4\right)}^2-18$:
    $\begin{array}{rcl}& & \\ 2{\left(x-4\right)}^2-18& =& 0\\ 2{\left(x-4\right)}^2& =& 18\\ {\left(x-4\right)}^2& =& 9\\ x-4& =& \pm \sqrt{9}\\ x-4& =& \pm 3\\ x& =& 4\pm 3\\ x& =& 1\mathrm{or}7\\ & & \end{array}$

  • The $x$-axis intercepts are at $\left(1,0\right)$ and $\left(7,0\right)$