Quadratic Functions

A full online textbook chapter on parabolas: axis of symmetry, vertex, roots, intercept form, vertex form, simultaneous equations and discriminant.

Online textbookIndependent studyWorked examplesRadford Mathematics

On this page

  1. Definition and shape
  2. Key features
  3. Axis and vertex
  4. Roots and intercepts
  5. Intercept form
  6. Vertex form
  7. Simultaneous equations
  8. Discriminant
  9. Videos
  10. Practice

1. What a quadratic function is

A quadratic function has the form

\[f(x)=ax^2+bx+c,\qquad a e0\]

Its graph is a parabola. In your notes, the key idea is that changing the coefficients changes the position and shape of the parabola, but a quadratic graph always has one line of symmetry and one turning point.

When \(a>0\)

The parabola opens upward and has a minimum point.

When \(a<0\)

The parabola opens downward and has a maximum point.

2. The features students must be able to find

When sketching or analysing a quadratic, do not simply draw a U-shape. Identify the mathematical features that control the sketch.

  • y-intercept: set \(x=0\). For \(ax^2+bx+c\), this is \((0,c)\).
  • x-intercepts / roots / zeros: set \(y=0\).
  • axis of symmetry: the vertical line through the vertex.
  • vertex: the maximum or minimum point.
  • concavity: whether the parabola opens up or down.

3. Axis of symmetry and vertex from standard form

For \(f(x)=ax^2+bx+c\), the axis of symmetry is

\[x=-\frac{b}{2a}\]

Once you have this \(x\)-value, substitute it into the function to find the vertex.

Worked example 1: find the axis and vertex

Find the axis of symmetry and vertex of \(y=2x^2-8x+5\).

Here \(a=2\), \(b=-8\) and \(c=5\).

\[x=-\frac{b}{2a}=-\frac{-8}{2(2)}=2\]

Now substitute \(x=2\):

\[y=2(2)^2-8(2)+5=8-16+5=-3\]

So the axis is \(x=2\) and the vertex is \((2,-3)\).

-112345-5-4-3-2-1123456789y0xx=2y=2x²−8x+5vertex (2,−3)(0,5)

The dashed line is the axis of symmetry. The vertex lies on this line.

4. Roots, zeros and x-intercepts

The roots, zeros and x-intercepts are all connected ideas. They are the values of \(x\) for which \(f(x)=0\).

Worked example 2: roots and intercepts

Sketch the key features of \(y=x^2-x-6\).

Factorise the quadratic:

\[x^2-x-6=(x-3)(x+2)\]

So the roots are \(x=3\) and \(x=-2\). The x-intercepts are \((3,0)\) and \((-2,0)\).

The y-intercept is found by setting \(x=0\), giving \((0,-6)\).

The axis of symmetry is halfway between the roots:

\[x=\frac{-2+3}{2}=\frac12\]
-4-3-2-112345-8-7-6-5-4-3-2-112345678y0xaxis x=1/2y=x²−x−6(−2,0)(3,0)vertex(0,−6)

The axis of symmetry is exactly halfway between the two roots.

5. Finding a parabola from its roots: intercept form

If the roots are known, use intercept form:

\[y=a(x-r_1)(x-r_2)\]

The roots give the brackets. One extra point is then used to find \(a\).

Worked example 3: equation from x-intercepts and a point

A parabola has roots \(-1\) and \(4\), and passes through \((0,-8)\). Find its equation.

Use intercept form:

\[y=a(x+1)(x-4)\]

Substitute the point \((0,-8)\):

\[-8=a(1)(-4)\]
\[a=2\]

Therefore

\[y=2(x+1)(x-4)\]
-3-2-1123456-12-11-10-9-8-7-6-5-4-3-2-1123456789101112y0xaxis x=1.5y=2(x+1)(x−4)(−1,0)(4,0)given point (0,−8)vertex

The graph crosses the x-axis at the two given roots. The point \((0,-8)\) fixes the value of \(a\).

6. Finding a parabola from its vertex and a point

If the vertex is known, use vertex form:

\[y=a(x-h)^2+k\]

The vertex is \((h,k)\). One extra point is used to find \(a\).

Worked example 4: equation from vertex and a point

A parabola has vertex \((3,-2)\) and passes through \((1,6)\). Find its equation.

Use vertex form with \((h,k)=(3,-2)\):

\[y=a(x-3)^2-2\]

Substitute \((1,6)\):

\[6=a(1-3)^2-2\]
\[6=4a-2\]
\[a=2\]

Therefore

\[y=2(x-3)^2-2\]
-11234567-4-3-2-1123456789101112y0xaxis x=3y=2(x−3)²−2vertex (3,−2)given point (1,6)symmetric point

The point \((1,6)\) is two units left of the axis, so the symmetric point \((5,6)\) is also on the parabola.

7. Finding a parabola using simultaneous equations

Sometimes you are given one of \(a\), \(b\) or \(c\), together with two points. In this case, substitute both points into \(y=ax^2+bx+c\) and solve the two equations.

If \(c\) is given, the y-intercept is known. If \(a\) is given, the concavity and vertical stretch are known. If \(b\) is given, the linear coefficient is known. The remaining two unknowns come from the two given points.

Worked example 5: equation when \(c\) is given

A quadratic has \(c=5\) and passes through \((1,10)\) and \((3,14)\). Find its equation.

Start with

\[y=ax^2+bx+5\]

Substitute \((1,10)\):

\[10=a+b+5\]
\[a+b=5\]

Substitute \((3,14)\):

\[14=9a+3b+5\]
\[3a+b=3\]

Now solve the simultaneous equations:

\[a+b=5\]
\[3a+b=3\]

Subtracting gives \(2a=-2\), so \(a=-1\). Then \(b=6\).

Therefore

\[y=-x^2+6x+5\]
-11234567-3-2-1123456789101112131415y0xaxis x=3y=−x²+6x+5c=5(1,10)(3,14)vertex

The graph passes through the two given points and has y-intercept \((0,5)\).

Mini example: equation when \(a\) is given

Suppose \(a=1\), and the parabola passes through \((1,2)\) and \((2,5)\).

Use \(y=x^2+bx+c\).

\[2=1+b+c\quad\Rightarrow\quad b+c=1\]
\[5=4+2b+c\quad\Rightarrow\quad 2b+c=1\]

Subtracting gives \(b=0\), so \(c=1\). The equation is \(y=x^2+1\).

Mini example: equation when \(b\) is given

Suppose \(b=-3\), and the parabola passes through \((0,2)\) and \((2,0)\).

Use \(y=ax^2-3x+c\). Since \((0,2)\) is on the graph, \(c=2\).

Substitute \((2,0)\):

\[0=4a-6+2\]
\[a=1\]

The equation is \(y=x^2-3x+2\).

8. Discriminant and number of roots

The discriminant tells you how many real roots a quadratic equation has.

\[\Delta=b^2-4ac\]
  • If \(\Delta>0\), there are two distinct real roots.
  • If \(\Delta=0\), there is one repeated real root.
  • If \(\Delta<0\), there are no real roots.

9. Related Radford Mathematics videos

Quadratic functions

Quadratic equations

Parabolas and roots

Worked example

10. Practice with solutions

  1. Find the axis and vertex of \(y=x^2-6x+11\).
    Show solution

    Axis: \(x=3\). Vertex: \((3,2)\).

  2. A parabola has roots \(2\) and \(5\), and passes through \((0,20)\). Find its equation.
    Show solution

    Use \(y=a(x-2)(x-5)\). Substitute \((0,20)\): \(20=a(-2)(-5)=10a\), so \(a=2\). Equation: \(y=2(x-2)(x-5)\).

  3. A parabola has vertex \((-1,4)\) and passes through \((1,0)\). Find its equation.
    Show solution

    Use \(y=a(x+1)^2+4\). Substitute \((1,0)\): \(0=4a+4\), so \(a=-1\). Equation: \(y=-(x+1)^2+4\).

  4. A quadratic has \(c=-2\) and passes through \((1,1)\) and \((2,8)\). Find its equation.
    Show solution

    Use \(y=ax^2+bx-2\). From \((1,1)\): \(a+b=3\). From \((2,8)\): \(4a+2b=10\), so \(2a+b=5\). Subtract to get \(a=2\), then \(b=1\). Equation: \(y=2x^2+x-2\).

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