Equation of a Quadratic Graph (SQA National 5 Maths): Revision Note

Determining the equation of a quadratic from its graph

How do I determine the equation of a quadratic in the form y = kx² from its graph?

• The graph of a quadratic with equation $\mathit{y}\mathbf{=}\mathit{k}{\mathit{x}}^{\mathbf{2}}$ will always have its turning point at the origin (0, 0)

  • $x=0$ will be its axis of symmetry

  • If $k$ is positive it will be an 'up' or $\cup$-shaped parabola

  • If $k$ is negative it will be a 'down' or $\cap$-shaped parabola

• If you are given a quadratic graph with its turning point at the origin and need to find its equation

  • The equation will be of the form $y=k{x}^2$

  • You just need to find the value of k by using the coordinates of another point on the graph

• For example, if a quadratic graph has its turning point at the origin and goes through the point $\left(2,-12\right)$

  • Substitute $x=2$ and $y=-12$ into $y=k{x}^2$

    • $-12=k{\left(2\right)}^2$

  • And solve to find the value of $k$

    • $-12=4k\Rightarrow k=-3$

  • The equation is $y=-3k^2$

How do I determine the equation of a quadratic in the form y = k(x+a)²+b from its graph?

• The graph of a quadratic with its equation in completed square form $\mathit{y}\mathbf{=}\mathit{k}{\mathbf{(}\mathit{x}\mathbf{+}\mathit{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}\mathit{b}$ will have its turning point at (-a, b)

  • $x=-a$ will be its axis of symmetry

  • If $k$ is positive it will be an 'up' or $\cup$-shaped parabola

  • If $k$ is negative it will be a 'down' or $\cap$-shaped parabola

• If you are given a quadratic graph with its turning point not at the origin and need to find its equation

  • The equation will be of the form $y=k{(x+a)}^2+b$

  • You can get the values of $a$ and $b$ from the coordinates of the turning point (-a, b)

    • Remember to switch the sign of the $x$-coordinate to get a

  • You can find the value of k by using the coordinates of another point on the graph

    • If that other point is the $\mathit{y}$-intercept remember that that point has an $x$-coordinate of 0

• For example, if a quadratic graph has its turning point at $\left(-2,3\right)$ and goes through the point $\left(1,21\right)$

  • From the turning point, $a=2$ and $b=3$

    • $y=k{\left(x+2\right)}^2+3$

  • From the point $\left(1,21\right)$, substitute $x=1$ and $y=21$ into $y=k{\left(x+2\right)}^2+3$

    • $21=k{\left(1+2\right)}^2+3$

  • And solve to find the value of $k$

    • $21=9k+3\Rightarrow 9k=18\Rightarrow k=2$

  • The equation is $y=2{\left(x+2\right)}^2+3$

• If you have found the values of $k,a$ and $b$in $y=k{(x+a)}^2+b$, and need to find the coordinates of the $\mathit{y}$-intercept

  • Remember that the $y$-intercept has coordinates $\left(0,c\right)$ for some value of $c$

  • Substitute $x=0$ into the equation and solve to find the value of $c$

  • For example, for $y=2{\left(x+2\right)}^2+3$

    • $y=2{\left(0+2\right)}^2+3=2\left(4\right)+3=8+3=11$

    • The $y$-intercept is at $\left(0,11\right)$