What do a, h and k do in vertex form?

In y equals a times x minus h all squared plus k, h shifts the parabola left or right, k shifts it up or down, and a stretches it, and reflects it when a is negative. The vertex is the point h, k.

What do a, h and k do in vertex form?

In y equals a times x minus h all squared plus k, h shifts the parabola left or right, k shifts it up or down, and a stretches it, and reflects it when a is negative. The vertex is the point h, k.

Year 11 Methods (Unit 1 & 2) Quadratics

Families Of Quadratic Polynomial Functions

Q: What is a family of quadratic functions?

All the parabolas obtained from y equals x squared by translations, dilations and reflections. Each member is described by y equals a times x minus h all squared plus k.

Q: How do I find the rule from the vertex and a point?

Write y equals a times x minus h all squared plus k with the vertex h, k, substitute the extra point, and solve for a.

Q: How do I find the rule from the x-intercepts?

Write y equals a times x minus x one, times x minus x two, using the intercepts, then substitute one more point to solve for a.

Q: How do I find the rule from three points?

Substitute each point into y equals a x squared plus b x plus c to get three equations, then solve them simultaneously for a, b and c.

Q: Why do I need an extra point beyond the vertex?

The vertex fixes the position but not the width. The extra point determines a, which selects the exact member of the family.

20 practice questions 1 video lesson Theory + worked examples

Understand the families of quadratic functions for Queensland Year 11 Mathematical Methods (QCAA). Every parabola is built from the squaring graph by shifting, stretching and reflecting it.

You will learn to read how each parameter shifts, stretches or reflects the curve, and to determine a parabola's rule from its turning point, from its x-intercepts, or from three points — the modelling skill used to fit a curve to data.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), a family of quadratics is all the parabolas built from \(y=x^2\) by translations, dilations and reflections. This page shows how the constants \(a\), \(h\) and \(k\) shape the graph in vertex form \(y=a(x-h)^2+k\), and how to determine the rule of a parabola from its vertex, its \(x\)-intercepts, or three points.

Every parabola belongs to the family of \(y=x^2\): it can be produced by shifting, stretching or flipping the basic parabola. The most useful form for describing a member is the vertex (turning-point) form \(y=a(x-h)^2+k\).

Here \((h,k)\) is the vertex. The constant \(h\) gives a horizontal translation, \(k\) a vertical translation, and \(a\) a dilation (vertical stretch). If \(a<0\) the parabola is also reflected in the \(x\)-axis, so it opens downwards.

The factored form \(y=a(x-x_1)(x-x_2)\) shows the \(x\)-intercepts \(x_1,x_2\) directly, while the general form \(y=ax^2+bx+c\) suits fitting through three points.

Match the form to the information. Vertex + a point → use \(y=a(x-h)^2+k\); two \(x\)-intercepts + a point → use \(y=a(x-x_1)(x-x_2)\); three general points → use \(y=ax^2+bx+c\).

Members of the \(y=x^2\) family, each translated to a new position.

Vertex form \(y=a(x-h)^2+k\): the vertex \((h,k)\) and axis \(x=h\) are read straight off.

Vertex (turning-point) form — vertex \((h,k)\):

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

y = a {(x - h)}^{2} + k

Factored form — \(x\)-intercepts \(x_1,x_2\):

\[y=a(x-x_1)(x-x_2)\]

y = a (x - x_{1}) (x - x_{2})

Find \(a\) with a known point. After placing the vertex or intercepts, substitute one more point on the curve and solve for \(a\); that single value fixes which member of the family it is.

How to determine the rule of a parabola

Choose the form that matches what you are given: vertex form for a turning point, factored form for \(x\)-intercepts, general form for three points.
Substitute the known feature (the vertex \((h,k)\) or the intercepts \(x_1,x_2\)) into that form.
Use a further point to solve for \(a\) (or solve the simultaneous equations for \(a,b,c\)), then write out the rule.

Example 1 — Transform the base parabola

The graph of \(y=x^2\) is reflected in the \(x\)-axis, then translated \(2\) units left and \(5\) units up. Write its rule.

Solution

Reflect in the \(x\)-axis — multiply by \(-1\):

\(y\)

\(=\)

\(-x^2\)

Translate \(2\) left — replace \(x\) with \(x+2\):

\(y\)

\(=\)

\(-(x+2)^2\)

Translate \(5\) up — add \(5\):

\(y\)

\(=\)

\(-(x+2)^2+5\)

Rule: \(y=-(x+2)^2+5\), a parabola with vertex \((-2,\,5)\) opening downwards.

y = - {(x + 2)}^{2} + 5

Example 2 — Rule from the vertex and a point

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

Solution

Use vertex form with \((h,k)=(1,-4)\):

\(y\)

\(=\)

\(a(x-1)^2-4\)

Substitute the point \((3,0)\) and solve for \(a\):

\(0\)	\(=\)	\(a(3-1)^2-4\)
\(0\)	\(=\)	\(4a-4\)
\(4a\)	\(=\)	\(4\)
\(a\)	\(=\)	\(1\)

Write the rule with \(a=1\):

\(y\)

\(=\)

\((x-1)^2-4\)

Rule: \(y=(x-1)^2-4\).

y = {(x - 1)}^{2} - 4

Example 3 — Rule from the intercepts and a point

A parabola cuts the \(x\)-axis at \(x=-2\) and \(x=4\) and passes through \((1,-9)\). Find its rule.

Solution

Use factored form with the intercepts:

\(y\)

\(=\)

\(a(x+2)(x-4)\)

Substitute \((1,-9)\) and solve for \(a\):

\(-9\)	\(=\)	\(a(1+2)(1-4)\)
\(-9\)	\(=\)	\(a(3)(-3)\)
\(-9\)	\(=\)	\(-9a\)
\(a\)	\(=\)	\(1\)

Write the rule with \(a=1\):

\(y\)

\(=\)

\((x+2)(x-4)\)

Rule: \(y=(x+2)(x-4)\), or expanded \(y=x^2-2x-8\).

y = (x + 2) (x - 4)

Example 4 — Rule from three points

Find the rule \(y=ax^2+bx+c\) of the parabola through \((0,3)\), \((1,2)\) and \((2,5)\).

Solution

Point \((0,3)\) gives \(c\) at once:

\(c\)

\(=\)

\(3\)

Point \((1,2)\): substitute \(x=1,y=2\):

\(a+b+c\)	\(=\)	\(2\)
\(a+b\)	\(=\)	\(-1\)

Point \((2,5)\): substitute \(x=2,y=5\):

\(4a+2b+c\)	\(=\)	\(5\)
\(4a+2b\)	\(=\)	\(2\)
\(2a+b\)	\(=\)	\(1\)

Subtract \(a+b=-1\) from \(2a+b=1\):

\(a\)

\(=\)

\(2\)

Back-substitute for \(b\):

\(2+b\)	\(=\)	\(-1\)
\(b\)	\(=\)	\(-3\)

Rule: \(y=2x^2-3x+3\).

y = 2 x^{2} - 3 x + 3

Common pitfalls

Sign of \(h\) in \((x-h)\). A vertex at \(x=-2\) gives \((x+2)^2\), because \(x-(-2)=x+2\). The bracket sign is the opposite of the \(x\)-coordinate.

Forgetting to find \(a\). The vertex or intercepts fix the shape's position but not its width; you still need one more point to pin down \(a\).

Reflection changes the sign of \(a\). Opening downwards means \(a<0\); do not leave \(a\) positive when the parabola is reflected.

Mixing up the forms. Use factored form only when you know the \(x\)-intercepts; use vertex form only when you know the turning point.

Frequently asked questions

What is a family of quadratic functions?

All the parabolas obtained from \(y=x^2\) by translations, dilations and reflections; each member is described by \(y=a(x-h)^2+k\).

What do a, h and k do in y = a(x-h)^2 + k?

\(h\) shifts left or right, \(k\) shifts up or down, and \(a\) stretches the parabola (and reflects it when \(a<0\)). The vertex is \((h,k)\).

How do I find the rule from the vertex and a point?

Write \(y=a(x-h)^2+k\) with the vertex \((h,k)\), substitute the extra point, and solve for \(a\).

How do I find the rule from the x-intercepts?

Write \(y=a(x-x_1)(x-x_2)\) using the intercepts, then substitute one more point to solve for \(a\).

How do I find the rule from three points?

Substitute each point into \(y=ax^2+bx+c\) to get three equations, then solve them simultaneously for \(a\), \(b\) and \(c\).

Why do I need an extra point beyond the vertex?

The vertex fixes the position but not the width; the extra point determines \(a\), which selects the exact member of the family.

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Families Of Quadratic Polynomial Functions

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How to determine the rule of a parabola

Common pitfalls

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Theory