The Graph Of y2=x
Understand the graph of y squared equals x for Queensland Year 11 Mathematical Methods (QCAA). This is a parabola lying on its side, opening to the right from the origin.
You will learn to plot it by choosing y-values, identify its vertex and axis of symmetry, and state its domain and range — and see why it is a relation rather than a function, since it fails the vertical line test.
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), the graph of \(y^2=x\) is a sideways parabola that opens to the right with its vertex at the origin. This page shows its shape, axis of symmetry, direction of opening, why it is a relation and not a function, and its domain and range.
The graph of \(y^2=x\) is a parabola lying on its side. Because \(x=y^2\) is never negative, the curve opens to the right, its vertex sits at the origin, and its axis of symmetry is the \(x\)-axis (\(y=0\)).
For each \(x>0\) there are two values of \(y\) (\(y=\pm\sqrt{x}\)), so a vertical line cuts the curve twice. It therefore fails the vertical-line test: \(y^2=x\) is a relation, not a function.
The general form \((y-k)^2=a(x-h)\) has vertex \((h,\,k)\) and axis of symmetry \(y=k\). When \(a>0\) it opens right; when \(a<0\) it opens left.
The basic curve and the single-valued form used for plotting:
The general sideways parabola, with vertex and axis:
How to analyse \((y-k)^2=a(x-h)\)
- Vertex and axis: read \((h,\,k)\) and write the axis of symmetry \(y=k\).
- Direction: if \(a>0\) it opens right, if \(a<0\) it opens left.
- Points: choose \(y\)-values, solve for \(x\); state the domain (\(x\ge h\) or \(x\le h\)) and range \(y\in\mathbb{R}\).
Rearrange — \(x=y^2\):
| \(x|_{y=3}\) | \(=\) | \((3)^2=9\) |
| \(x|_{y=-2}\) | \(=\) | \((-2)^2=4\) |
So the curve passes through \((9,\,3)\) and \((4,\,-2)\).
Shape — \(x=y^2\ge 0\):
| \(\text{opens}\) | \(=\) | \(\text{right}\) |
| \(\text{vertex}\) | \(=\) | \((0,\,0)\) |
\((9,\,3)\) and \((4,\,-2)\); opens right, vertex \((0,\,0)\), axis \(y=0\).
Substitute \(x=1\):
| \(y^2\) | \(=\) | \(4(1)\) |
| \(y^2\) | \(=\) | \(4\) |
| \(y\) | \(=\) | \(\pm 2\) |
So the curve meets \(x=1\) at \((1,\,2)\) and \((1,\,-2)\).
One \(x\)-value gives two \(y\)-values, so a vertical line cuts the curve twice: it fails the vertical-line test and is a relation, not a function.
Meets \(x=1\) at \((1,\,2)\) and \((1,\,-2)\); it is a relation, not a function.
Match to \((y-k)^2=a(x-h)\):
| \((y-1)^2\) | \(=\) | \(1\cdot(x-(-2))\) |
| \(h\) | \(=\) | \(-2\) |
| \(k\) | \(=\) | \(1\) |
| \(a\) | \(=\) | \(1\) |
Vertex \((-2,\,1)\), axis \(y=1\); since \(a=1>0\) it opens right.
Point — put \(y=3\):
| \((3-1)^2\) | \(=\) | \(x+2\) |
| \(4\) | \(=\) | \(x+2\) |
| \(x\) | \(=\) | \(2\) |
Vertex \((-2,\,1)\), axis \(y=1\), opens right; passes through \((2,\,3)\).
Direction — the coefficient of \(x\) is negative:
| \(x\) | \(=\) | \(-\dfrac{y^2}{2}\) |
| \(\text{opens}\) | \(=\) | \(\text{left}\) |
Point — put \(y=4\):
| \((4)^2\) | \(=\) | \(-2x\) |
| \(16\) | \(=\) | \(-2x\) |
| \(x\) | \(=\) | \(-8\) |
Since \(x=-\tfrac{y^2}{2}\le 0\), the domain is \(x\le 0\) and the range is \(y\in\mathbb{R}\).
Opens left; domain \(x\le 0\), range \(y\in\mathbb{R}\); passes through \((-8,\,4)\).
Common pitfalls
Frequently asked questions
What does the graph of y squared equals x look like?
It is a parabola lying on its side, opening to the right, with its vertex at the origin and the \(x\)-axis as its axis of symmetry.
Is y squared equals x a function?
No. Each positive \(x\) gives two \(y\)-values (\(y=\pm\sqrt{x}\)), so it fails the vertical-line test and is a relation, not a function.
What are the domain and range of y squared equals x?
The domain is \(x\ge 0\) and the range is all real \(y\), since \(x=y^2\) can never be negative but \(y\) can be anything.
How do you plot points on y squared equals x?
Choose values of \(y\) and compute \(x=y^2\); for example \(y=2\) gives \(x=4\), so \((4,\,2)\) is on the curve.
Which way does the sideways parabola open?
For \((y-k)^2=a(x-h)\) it opens right when \(a>0\) and left when \(a<0\).