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 $k > 0$.
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