Vertex form: how a, h and k move a parabola

Transformations

What do a, h and k each do to y = a(x − h)² + k?

Stretch, flip and slide the parabola y = x² with three sliders. See where the vertex goes, why x − 2 moves the graph right, and when the roots appear.

Drag the ringed points on the board.

Middle school
High school
Vertex form
Translations
Axis of symmetry
Roots and discriminant

a

opens down

opens up

1.0

h

0.0

k

0.0

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

$a$: 1.0
$\text{vertex}\ x$: 0.0
$\text{vertex}\ y$: 0.0
$x_1$: 0.00
$x_2$: 0.00

Meets the x-axisreal roots

Guided tour

1The parent graph
With $a=1$ , $h=0$ , $k=0$ the curve is $y=x^2$ : vertex at the origin, opening up, one root at $x=0$ .
2a stretches and flips
At $a=2$ every height doubles, so the parabola looks narrower. At $a=-0.5$ heights are halved and flipped: it opens downward.
3h slides sideways, against its sign
Set $h=2$ : the formula reads $(x-2)^2$ and the vertex moves to $x=2$ , to the right. The bracket is zero when $x=2$ , which is where the vertex now sits.
4k lifts, and roots appear
Now $k=-3$ drops the vertex to $(2,-3)$ and the curve crosses the $x$ -axis twice, at $x=2\pm\sqrt3$ . Raise $k$ above $0$ and the roots disappear.

Explanation

1
Reading the vertex form
$y=a(x-h)^2+k$
The square $(x-h)^2$ is never negative and is $0$ only at $x=h$ . So the turning point, the vertex, is $(h,k)$ , and the vertical line $x=h$ is the axis of symmetry.
2
What a does
$a$ multiplies every height measured from the vertex. $|a|>1$ stretches the curve (narrower), $0<|a|<1$ squashes it (wider), and $a<0$ flips it so it opens downward. At $a=0$ the square vanishes and the graph is the flat line $y=k$ .
3
Why x − h moves right
The new graph at $x$ copies the old graph at $x-h$ . To reach the old vertex at $0$ you need $x-h=0$ , that is $x=h$ . So subtracting $h$ inside moves the graph right by $h$ , while adding $k$ outside moves it up by $k$ . Inside changes act on $x$ in the opposite direction; outside changes act on $y$ directly.
4
Roots from the vertex form
Set $y=0$ : $(x-h)^2=-\frac{k}{a}$ . Real roots exist exactly when $-\frac{k}{a}\ge0$ , that is when the vertex and the opening point to opposite sides of the $x$ -axis:
$x=h\pm\sqrt{-\frac{k}{a}}$
Expanding to $y=ax^2+bx+c$ gives $h=-\frac{b}{2a}$ , and the condition becomes the familiar discriminant test $b^2-4ac\ge0$ .

Check yourself

Q01

What is the vertex of $y=3(x+1)^2-4$ ?

Q02

How many real roots does $y=-2(x-3)^2-1$ have?

Q03

Which transformation turns $y=x^2$ into $y=(x-5)^2$ ?

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