Sketching Translations of y=x^2
Every question with a fully worked solution.
- Sketching Translations of y=x^2 - Video - Example translating parabola Watch
Theory
Adding or subtracting a number moves the graph of \(y = x^2\) without changing its shape. This Year 9 maths page shows how \(y = x^2 + k\) moves the parabola up or down, how \(y = (x - h)^2\) moves it left or right, and how to read the turning point \((h, k)\) straight from \(y = a(x - h)^2 + k\). It also covers finding the intercepts, sketching a translated parabola and finding its equation from a graph or a description.
A translation slides a graph up, down, left or right without changing its shape. A reflection in the \(x\)-axis flips it upside down, and a dilation makes it narrower or wider (see the page on \(y = ax^2\)).
Adding \(k\) outside the square, \(y = x^2 + k\), moves the graph up \(k\) units (down if \(k\) is negative). Replacing \(x\) with \(x - h\), \(y = (x - h)^2\), moves it right \(h\) units, so \(x + 2\) moves it \(2\) units left.
The turning point form of a parabola is \(y = a(x - h)^2 + k\). Its turning point (vertex) is \((h, k)\) and its axis of symmetry is \(x = h\). If \(a < 0\) the parabola is reflected in the \(x\)-axis and is concave down.
In a graphing app, add sliders for \(h\) and \(k\) in \(y = (x - h)^2 + k\). Dragging \(k\) slides the curve up and down; dragging \(h\) slides it sideways, and the turning point always sits at \((h, k)\).
Turning point form, and the axis of symmetry of \(y = ax^2 + bx + c\):
| Equation | Move from the graph of \(y = f(x)\) |
|---|---|
| \(y = f(x) + k\) | up \(k\) |
| \(y = f(x) - k\) | down \(k\) |
| \(y = f(x - h)\) | right \(h\) |
| \(y = f(x + h)\) | left \(h\) |
| \(y = -f(x)\) | reflect in the \(x\)-axis |
How to sketch \(y = a(x - h)^2 + k\)
- Shape: \(a > 0\) concave up, \(a < 0\) concave down.
- Turning point: \((h, k)\). Watch the sign of \(h\): \((x + 1)^2\) means \(h = -1\).
- \(y\)-intercept: put \(x = 0\).
- \(x\)-intercepts: put \(y = 0\), get the square on its own and take the square root of both sides (\(\pm\)). If the square would have to be negative, there are none.
- Plot these points and draw a smooth, symmetric curve.
Writing the equation
- From a description: right \(h\) gives \(x - h\) in the bracket, up \(k\) gives \(+\,k\) at the end. A reflection in the \(x\)-axis puts a minus sign in front of the bracket.
- From a graph: read the turning point \((h, k)\), write \(y = a(x - h)^2 + k\), then substitute another point to find \(a\). If the graph is just \(y = x^2\) moved, \(a = 1\).
From \(y = ax^2 + bx + c\)
- Axis: \(x = -\dfrac{b}{2a}\).
- Turning point: substitute the axis value to find \(y\).
- Intercepts: \(y\)-intercept \(c\); factorise to find the \(x\)-intercepts.
(i) \(y = (x + 3)^2 - 5\)
(ii) \(y = 2 - (x - 4)^2\)
Compare with \(y = a(x - h)^2 + k\). The turning point is \((h, k)\).
(i) \(x + 3 = x - (-3)\), so \(h = -3\) and \(k = -5\). The graph moves \(3\) left and \(5\) down.
Turning point \((-3, -5)\), a minimum.
(ii) Rewrite as \(y = -(x - 4)^2 + 2\), so \(a = -1\), \(h = 4\) and \(k = 2\). The graph is reflected in the \(x\)-axis, then moved \(4\) right and \(2\) up.
Turning point \((4, 2)\), a maximum.
\(x + 2 = x - (-2)\), so the turning point is \((-2, -4)\); \(a = 1 > 0\), so concave up.
\(y\)-intercept: put \(x = 0\).
| \(y\) | \(=\) | \((0 + 2)^2 - 4\) |
| \(\) | \(=\) | \(\boldsymbol{0}\) |
\(x\)-intercepts: put \(y = 0\).
| \((x + 2)^2\) | \(=\) | \(4\) |
| \(x + 2\) | \(=\) | \(\pm 2\) |
| \(x\) | \(=\) | \(\boldsymbol{0} \text{ or } \boldsymbol{-4}\) |
(i) the axis of symmetry
(ii) the turning point
(iii) the intercepts.
(i) Use \(x = -\dfrac{b}{2a}\) with \(a = 1\), \(b = 6\).
| \(x\) | \(=\) | \(-\dfrac{6}{2 \times 1} = \boldsymbol{-3}\) |
(ii) Substitute \(x = -3\):
| \(y\) | \(=\) | \((-3)^2 + 6(-3) + 8\) |
| \(\) | \(=\) | \(9 - 18 + 8\) |
| \(\) | \(=\) | \(-1\) |
The turning point is \((-3, -1)\).
(iii) \(y\)-intercept: \(x = 0\) gives \(y = \mathbf{8}\). For the \(x\)-intercepts, factorise:
| \((x + 2)(x + 4)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{-4} \text{ or } \boldsymbol{-2}\) |
In turning point form:
| \(y\) | \(=\) | \((x + 3)^2 - 1\) |
It is the graph of \(y = x^2\) moved \(3\) left and \(1\) down.
Put the turning point in: \(h = -1\), \(k = 5\).
| \(y\) | \(=\) | \(a(x + 1)^2 + 5\) |
Substitute the point \((1, -3)\) to find \(a\):
| \(-3\) | \(=\) | \(a(1 + 1)^2 + 5\) |
| \(-8\) | \(=\) | \(4a\) |
| \(a\) | \(=\) | \(-2\) |
| \(y\) | \(=\) | \(\boldsymbol{-2(x + 1)^2 + 5}\) |
\(a\) is negative, so the parabola is concave down, which fits: \((1, -3)\) is below the turning point.
Common pitfalls
Frequently asked questions
What is turning point form?
It is \(y = a(x - h)^2 + k\). The turning point is \((h, k)\), the axis of symmetry is \(x = h\), and the sign of \(a\) tells you if the parabola is concave up or down.
Why does \((x + 6)^2\) move the graph left?
The bracket is zero when \(x = -6\), so the turning point is now at \(x = -6\). Everything that happened at \(x = 0\) now happens \(6\) units earlier, to the left.
How do you find the turning point of \(y = ax^2 + bx + c\)?
Find the axis of symmetry with \(x = -\dfrac{b}{2a}\), then substitute that \(x\)-value into the equation to find the \(y\)-coordinate.
How do I find the equation of a parabola from its graph?
Read the turning point \((h, k)\) and write \(y = a(x - h)^2 + k\). Then substitute another point on the graph, such as an intercept, to find \(a\).
What does \(f(x - h) + k\) mean?
It is the graph of \(y = f(x)\) moved \(h\) units to the right and \(k\) units up.