Vertex and Axis of Symmetry of a Parabola
Vertex form makes the vertex and the axis of symmetry easy to read — as long as you handle the sign correctly.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
For $f(x) = a(x - h)^2 + k$, the vertex is $(h, k)$ and the axis of symmetry is the vertical line $x = h$. Note the minus sign: $(x + 4)^2$ means $h = -4$.
How to solve one, step by step
Example: $f(x) = 2(x - 4)^2 + 1$.
- Match to $a(x - h)^2 + k$: $h = 4$, $k = 1$.
- Vertex $(4, 1)$; axis of symmetry $x = 4$.
The mistakes that cost points
- Not flipping the sign of h. $(x + 4)^2$ gives $h = -4$, not $+4$.
- Confusing the vertex x with the axis value. The axis of symmetry is $x = h$, the same value as the vertex's $x$-coordinate — don't mix it with $k$.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
The graph of $y = f(x)$ is a parabola, where $f(x) = 2(x - 6)^2 - 3$. Which of the following is the vertex of this parabola?
- A$(6, 3)$
- B$(-6, -3)$
- C$(6, -3)$
- D$(-6, 3)$
Show solution
Answer: C, $(6, -3)$. In vertex form $a(x-h)^2 + k$, the vertex is $(h, k)$. The expression $(x - 6)$ gives $h = 6$. The constant is $-3$, so $k = -3$. The vertex is $(6, -3)$.
Medium
The function $f(x) = 2(x - k)^2 + 3$ has its axis of symmetry at $x = 5$. What is the value of $k$?
- A$10$
- B$-5$
- C$5$
- D$3$
Show solution
Answer: C, $5$. In vertex form $f(x) = a(x - h)^2 + k_{vertex}$, the axis of symmetry is $x = h$. Here $h = k$, so the axis of symmetry is $x = k$. Setting $k = 5$ gives axis of symmetry $x = 5$ ✓.
Hard
For which values of $k$ will the parabola $-16(x+k) = (y+1)^2$ have no real $y$-intercept?
- A$k > 0$
- B$k < 0$
- C$k \geq 0$
- DThe parabola will always have a real $y$-intercept, regardless of the value of $k$.
Show solution
Answer: A, $k > 0$. To find the $y$-intercept(s), set $x = 0$: $-16(0+k) = (y+1)^2$, which gives $-16k = (y+1)^2$. For real $y$ to exist, we need $(y+1)^2 \geq 0$, which means $-16k \geq 0$, which means $k \leq 0$. The parabola has no real $y$-intercept when this fails — that is, when $-16k < 0$, which means $k > 0$.
Find your exact gaps
Take a free, full-length practice test and see precisely which question types trip you up.
Practice this for freeNo credit card. Register once, take as many tests as you like.