Quadratic Graphs: Roots and Turning Points

Reading a quadratic graph

A quadratic function can be written as $y=ax^2+bx+c$, where $a\ne0$. Its graph is a smooth, symmetrical curve called a parabola. If $a>0$, it opens upwards and has a minimum; if $a<0$, it opens downwards and has a maximum. These features, together with the intercepts, let you make an accurate sketch without plotting a long table of values.

The $y$-intercept is where $x=0$, so it is $(0,c)$. At an $x$-intercept, $y=0$. Its $x$-coordinate is a root of $ax^2+bx+c=0$. Thus finding roots algebraically also tells you where the graph meets the $x$-axis.

Roots and symmetry

Consider $y=x^2-4x+3$. Setting $y=0$ and factorising gives $(x-1)(x-3)=0$, so the roots are $x=1$ and $x=3$. The $x$-intercepts are therefore $(1,0)$ and $(3,0)$; the $y$-intercept is $(0,3)$.

The two roots are equally far from the vertical line of symmetry. Its $x$-coordinate is their midpoint, $(1+3)/2=2$. Substituting $x=2$ into the function gives $y=-1$, so the turning point is $(2,-1)$. Since the coefficient of $x^2$ is positive, this is a minimum. You can now sketch a smooth upward-opening curve through the intercepts, mirrored about $x=2$.

This midpoint method depends on having two real roots. A quadratic may instead touch the $x$-axis at its turning point, or miss it altogether. Completing the square locates the turning point in all three cases.

Finding the turning point by completing the square

Completing the square rewrites the quadratic as $y=a(x-h)^2+k$. Because $(x-h)^2$ is smallest when $x=h$, the turning point is $(h,k)$ and the line of symmetry is $x=h$. It is a minimum when $a>0$ and a maximum when $a<0$.

To see where this form comes from, expand $(x+\frac b2)^2=x^2+bx+(\frac b2)^2$. The square contains an extra $(\frac b2)^2$, so subtract that amount to keep the original expression unchanged:

$x^2+bx+c=\left(x+\frac b2\right)^2+c-\left(\frac b2\right)^2.$

For the earlier graph, $x^2-4x+3=(x-2)^2-1$. This confirms its turning point $(2,-1)$ without first finding the roots. Setting $y=0$ also gives $(x-2)^2=1$, hence $x-2=\pm1$ and the same two roots, $1$ and $3$.

If the coefficient of $x^2$ is not $1$, first factor it out of the quadratic and linear terms, then complete the square inside the brackets. For example,

$2x^2+8x+10=2(x^2+4x+5)=2\big((x+2)^2+1\big)=2(x+2)^2+2.$

This graph has a minimum at $(-2,2)$ and symmetry line $x=-2$. Its minimum $y$-value is $2$, so it has no real roots: the curve stays above the $x$-axis. Its $y$-intercept is $(0,10)$. These features are enough to sketch its position and shape.

Features for a sketch

For $y=ax^2+bx+c$, the $y$-intercept is $(0,c)$. Solve $ax^2+bx+c=0$ algebraically for the roots; each real root gives an $x$-intercept. The graph opens upwards if $a>0$ and downwards if $a<0$.

Turning point

With two real roots $r_1,r_2$, the symmetry line is $x=(r_1+r_2)/2$; substitute this value to find the turning point’s $y$-coordinate. Alternatively, complete the square: $y=a(x-h)^2+k$ has turning point $(h,k)$ and symmetry line $x=h$. This method also works when there are fewer than two real roots.