Quadratic graphs 

​​In a nutshell

A quadratic graph $$y=ax^2+bx+c$$ has a parabolic shape, meaning it is either $$\cup$$-shaped or $$\cap$$-shaped. These graphs can be drawn once some key features have been identified, including the shape, any axes intercepts and the location of the turning point.

Notation

Often graphs are given with function notation:

​$$\boxed{f(x)=ax^2+bx+c}$$​​

The equation for the graph is then given as 

​$$\boxed{y=f(x)}$$​​

This is the same as saying $$y=ax^2+bx+c$$.

Note: $$a$$, $$b$$ and $$c$$ are constants. 

Shape

The shape of a quadratic curve is determined by the value of $$a$$ in the quadratic equation. If $$a>0$$, then the curve is $$\cup$$​-shaped (below left) and is called a positive quadratic, and if $$a<0$$, then it is $$\cap$$-shaped (below right) and is called a negative quadratic.

Note: If $$a=0$$ then you don't have a quadratic.

Intercepts

​​$$y$$-intercept​

Quadratic graphs will always intercept the $$y$$-axis once. To find this point, insert $$x=0$$ (since the $$y$$-axis is the line along which $$x$$ is always zero) into the equation $$y=ax^2+bx+c$$. This will give $$y=c$$. Hence, the point $$(0,c)$$ is the $$y$$-intercept. 

​​$$x$$-intercept(s)

There are either zero, one or two $$x$$-intercepts (points where the curve intersects with the $$x$$-axis) of a quadratic graph. Since the $$x$$-axis is the line along which $$y$$ is zero, insert $$y=0$$ into the quadratic equation $$y=ax^2+bx+c$$. Hence to find the $$x$$-intercept(s), you must solve 

​$$0=ax^2+bx+c$$​​

This is a case of either factorising, using the quadratic formula or completing the square. If there are two solutions, then there are two $$x$$-intercepts and your quadratic graph will look similar to the two above. If there is one solution to this equation, then there is one $$x$$-intercept - this means that the curve touches the $$x$$-axis at one point, for example the graph below:

In particular, the $$x$$-intercept for this curve is at $$(3,0)$$. If the quadratic equation has no solutions, then it does not touch the $$x$$-axis, so it is either entirely above or entirely below the $$x$$-axis, for example, either of the two graphs on the grid below:

If there are no $$x$$-intercepts, then you have to rely on the shape and the turning point to sketch the graph.​

Example 1

Find the shape and the intercept(s) of the graph $$y=f(x)$$ where

​$$f(x)=x^2+x-12$$​​

The shape is given by the positivity of $$a$$. Here $$a=1$$, so the quadratic is positive (and hence $$\underline{\cup-\textit{shaped}}$$).

The $$y$$-intercept is given by $$c$$, which is $$-12$$, so the $$y$$-intercept is at $$\underline{(0,-12)}$$.

Any $$x$$-intercepts are found by solving for $$y=0$$, in other words, solving $$0=x^2+x-12$$. By factorising, you find

$$0=(x-3)(x+4)$$​​

Hence the $$x$$-intercepts are when $$x=3$$ and $$x=-4$$. So the coordinates of those points are $$\underline{(3,0)}$$ and $$\underline{(-4,0)}$$.

​​Turning point

Quadratics have exactly one turning point. This is the point where the graph "turns around". For positive quadratics, this is the minimum point, for the negative quadratics, this is the maximum point. This point can be found by completing the square on the quadratic. Once the quadratic is in the form $$y=A(x+B)^2+C$$, the turning point is at $$(-B,C)$$.

Example 2

By finding the key features (the shape, the intercept(s), the turning point), sketch the graph $$y=f(x)$$ where 

$$f(x)=-x^2+2x+3$$​​

The value of $$a$$ is negative, so the curve is $$\cap$$-shaped. Since $$c=3$$, the $$y$$-intercept is at $$(0,3)$$. To find the $$x$$-intercepts, set $$y$$ to zero and solve:

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

​​​

Hence $$x=3$$ and $$x=-1$$. So there are two $$x$$-intercepts at $$(3,0)$$ and $$(-1,0)$$. Finally, the turning point can be found by completing the square:

$$\begin{aligned}y&=-x^2+2x+3\\\space\\y&=-(x^2-2x)+3\\\space\\y&=-((x-1)^2-1)+3\\\space\\y&=-(x-1)^2+1+3\\\space\\y&=-(x-1)^2+4 \end{aligned}$$​​

So the turning point is at $$(1,4)$$. A sketch of this curve is given below: