The Graph Of y=SQRT{X}
Understand the graph of y equals the square root of x for Queensland Year 11 Mathematical Methods (QCAA). It is a half-parabola that starts at an endpoint and rises to the right.
You will learn to find the endpoint and intercepts, state the domain and range, describe how it behaves for large x, and see the effect of the parameters when a negative coefficient reflects the curve so it falls.
Every question with a fully worked solution.
- The Graph Of y=SQRT{X} - Video - The square root graph Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), the graph of \(y=\sqrt{x}\) is a half-parabola that begins at the origin and rises to the right. This page shows how to find the endpoint, domain and range, intercepts, and how the coefficient \(a\) in \(y=a\sqrt{x-h}+k\) stretches or reflects the curve.
The graph of \(y=\sqrt{x}\) is the top half of a sideways parabola. It starts at the endpoint \((0,\,0)\) and increases to the right; it is defined only for \(x\ge 0\) because you cannot take the square root of a negative number.
The general form is \(y=a\sqrt{x-h}+k\). The endpoint is \((h,\,k)\); the domain is \(x\ge h\). The coefficient \(a\) is a dilation: if \(a>0\) the curve rises (range \(y\ge k\)); if \(a<0\) it is reflected and falls (range \(y\le k\)).
The general square-root function, its endpoint and domain:
Range depends on the sign of \(a\):
How to analyse \(y=a\sqrt{x-h}+k\)
- Endpoint: read \((h,\,k)\) directly from the rule.
- Direction and range: \(a>0\) rises (range \(y\ge k\)); \(a<0\) falls (range \(y\le k\)); the domain is \(x\ge h\).
- Intercepts: substitute \(x=0\) for the \(y\)-intercept, and set \(y=0\) then square to find the \(x\)-intercept.
Values — take each square root:
| \(\sqrt{0}\) | \(=\) | \(0\) |
| \(\sqrt{1}\) | \(=\) | \(1\) |
| \(\sqrt{4}\) | \(=\) | \(2\) |
| \(\sqrt{9}\) | \(=\) | \(3\) |
The endpoint is \((0,\,0)\) and the curve rises to the right.
Domain and range — \(\sqrt{x}\) needs \(x\ge 0\):
| \(\text{domain}\) | \(=\) | \(x\ge 0\) |
| \(\text{range}\) | \(=\) | \(y\ge 0\) |
Passes through \((0,0),(1,1),(4,2),(9,3)\); domain \(x\ge 0\), range \(y\ge 0\).
Endpoint — read \(h\) and \(k\):
| \(h\) | \(=\) | \(2\) |
| \(k\) | \(=\) | \(1\) |
Endpoint \((2,\,1)\); since \(a=1>0\) the domain is \(x\ge 2\) and range \(y\ge 1\).
Point — substitute \(x=6\):
| \(y\) | \(=\) | \(\sqrt{6-2}+1\) |
| \(=\) | \(\sqrt{4}+1\) | |
| \(=\) | \(3\) |
Endpoint \((2,\,1)\); domain \(x\ge 2\), range \(y\ge 1\); passes through \((6,\,3)\).
Endpoint:
| \((h,\,k)\) | \(=\) | \((0,\,-3)\) |
\(y\)-intercept — put \(x=0\):
| \(y\) | \(=\) | \(2\sqrt{0}-3\) |
| \(=\) | \(-3\) |
\(x\)-intercept — put \(y=0\):
| \(0\) | \(=\) | \(2\sqrt{x}-3\) |
| \(2\sqrt{x}\) | \(=\) | \(3\) |
| \(\sqrt{x}\) | \(=\) | \(\dfrac{3}{2}\) |
| \(x\) | \(=\) | \(\dfrac{9}{4}\) |
Endpoint \((0,\,-3)\); intercepts \((0,\,-3)\) and \(\left(\tfrac{9}{4},\,0\right)\); range \(y\ge -3\).
Endpoint and direction:
| \((h,\,k)\) | \(=\) | \((-1,\,2)\) |
| \(a\) | \(=\) | \(-1\) |
Since \(a<0\) the curve falls: domain \(x\ge -1\), range \(y\le 2\).
\(y\)-intercept — put \(x=0\):
| \(y\) | \(=\) | \(-\sqrt{0+1}+2\) |
| \(=\) | \(-1+2\) | |
| \(=\) | \(1\) |
\(x\)-intercept — put \(y=0\):
| \(0\) | \(=\) | \(-\sqrt{x+1}+2\) |
| \(\sqrt{x+1}\) | \(=\) | \(2\) |
| \(x+1\) | \(=\) | \(4\) |
| \(x\) | \(=\) | \(3\) |
Endpoint \((-1,\,2)\), falls; range \(y\le 2\); intercepts \((0,\,1)\) and \((3,\,0)\).
Common pitfalls
Frequently asked questions
What does the graph of y equals square root of x look like?
It is a half-parabola: it starts at the origin and rises to the right, defined only for \(x\ge 0\).
What is the endpoint of y equals a root x minus h plus k?
The endpoint is \((h,\,k)\); it is where the curve begins, since the expression under the root is zero there.
What are the domain and range of a square-root graph?
The domain is \(x\ge h\). The range is \(y\ge k\) when \(a>0\), and \(y\le k\) when \(a<0\).
How does a negative coefficient a change the graph?
A negative \(a\) reflects the curve in the horizontal line through the endpoint, so it falls to the right instead of rising.
How do you find the x-intercept of a square-root graph?
Set \(y=0\), isolate the square root, then square both sides and solve for \(x\).