Big Idea
A quadratic function models a situation where the graph bends instead of staying straight. Its graph is a parabola.
One of the most useful ways to write a quadratic is vertex form:
$$y = a(x-h)^2 + k$$
This form tells us the graph's most important features right away:
- • where the graph turns
- • whether it opens up or down
- • whether it has a maximum or minimum value
1. Key Definitions
Quadratic function
A function whose highest power of $x$ is $2$.
Parabola
The U-shaped graph of a quadratic function.
Vertex
The turning point of the parabola. It is the highest point if the graph opens down, and the lowest point if the graph opens up.
Axis of symmetry
The vertical line that goes through the vertex and splits the parabola into two matching halves.
Maximum or minimum
- • If $a > 0$, the parabola opens upward, so it has a minimum
- • If $a < 0$, the parabola opens downward, so it has a maximum
2. Core Rules and Properties
For a quadratic in vertex form, $y = a(x-h)^2 + k$, these are the key facts:
Vertex
$(h,\, k)$
Axis of symmetry
$x = h$
Opens upward
when $a > 0$
Opens downward
when $a < 0$
Narrower than $y = x^2$
when $|a| > 1$
Wider than $y = x^2$
when $0 < |a| < 1$
Watch the sign inside the parentheses!
- • $x - h$ shifts the graph right $h$ units
- • $x + h$ shifts the graph left $h$ units
The sign inside the parentheses often tricks people — always check it carefully.
3. Example: Reading the Equation
Find the vertex, axis of symmetry, opening direction, and maximum or minimum value of
$$f(x) = -2(x+3)^2 + 5$$
Solution
Compare to $y = a(x-h)^2 + k$. Rewrite the inside first:
$$x + 3 = x - (-3) \quad \Rightarrow \quad h = -3$$
So we have: $a = -2$, $h = -3$, $k = 5$
Vertex
$(-3,\; 5)$
Axis of symmetry
$x = -3$
Direction
Opens downward since $a = -2 < 0$
Maximum value
$5$, reached at $(-3,\; 5)$
The graph reaches its highest point at $(-3,\; 5)$.
4. Interpreting the Equation in Words
Vertex form lets you describe the graph without plotting many points. For example, in
$$y = \tfrac{1}{2}(x-1)^2 - 4$$
you can say:
- • the graph starts from the basic parabola $y = x^2$
- • it shifts right 1
- • it shifts down 4
- • it opens upward
- • it is wider than $y = x^2$ because $\tfrac{1}{2}$ is between $0$ and $1$
That already tells a lot about the graph — before plotting a single point.
5. Example: Describing and Finding Points
Describe the graph of
$$g(x) = \tfrac{1}{2}(x-1)^2 - 4$$
and find two points on the graph besides the vertex.
Solution
Step 1 — Identify the main features:
Vertex
$(1,\; -4)$
Axis of symmetry
$x = 1$
Direction
Opens upward since $a = \tfrac{1}{2} > 0$
Shape
Wider than $y = x^2$ since $0 < \tfrac{1}{2} < 1$
Step 2 — Find two points equally spaced from $x = 1$.
Choose $x = 3$ and $x = -1$. Each is 2 units from $x = 1$.
For $x = 3$:
$$g(3) = \tfrac{1}{2}(3-1)^2 - 4 = \tfrac{1}{2}(2)^2 - 4 = \tfrac{1}{2}(4) - 4 = 2 - 4 = -2$$
Point: $(3,\; -2)$
For $x = -1$:
$$g(-1) = \tfrac{1}{2}(-1-1)^2 - 4 = \tfrac{1}{2}(-2)^2 - 4 = \tfrac{1}{2}(4) - 4 = 2 - 4 = -2$$
Point: $(-1,\; -2)$
Both points give $y = -2$. That makes sense — the graph is symmetric around $x = 1$, so equal distances left and right always give the same $y$-value.
6. Writing a Quadratic from a Vertex and Point
Sometimes you know the vertex and one other point. That is enough to build the equation.
Method
- Start with vertex form: $y = a(x-h)^2 + k$
- Plug in $(h, k)$ from the vertex
- Substitute the other point $(x, y)$ to solve for $a$
Write the equation of the quadratic with vertex $(4,\; -1)$ that passes through $(6,\; 7)$.
Solution
Since the vertex is $(4, -1)$, start with:
$$y = a(x-4)^2 - 1$$
Substitute the point $(6,\; 7)$:
$$7 = a(6-4)^2 - 1$$
$$7 = a(2)^2 - 1$$
$$7 = 4a - 1$$
Add $1$ to both sides:
$$8 = 4a$$
Divide by $4$:
$$a = 2$$
The equation is:
$$y = 2(x-4)^2 - 1$$
7. Mental Checks and Estimation
Before doing a lot of work, a quick mental check saves time and catches errors early.
Look at the sign of $a$ first
That tells you whether the graph goes up or down.
Look at the vertex
That tells you where the turning point is.
Check symmetry
Points the same distance left and right of the axis should have the same $y$-value.
Estimate intercepts
If the vertex is below the $x$-axis and the graph opens upward, it will usually cross the $x$-axis twice.
8. Example: Finding x-Intercepts
Find the $x$-intercepts of
$$y = (x-5)^2 - 4$$
Then explain why the answers make sense before solving.
Solution
Mental check first:
The vertex is $(5,\; -4)$. The graph opens upward (coefficient of the squared term is positive). Since the lowest point is below the $x$-axis, the parabola should cross the $x$-axis two times, and both intercepts should be equally spaced around $x = 5$.
Now solve — set $y = 0$:
$$0 = (x-5)^2 - 4$$
Add $4$ to both sides:
$$4 = (x-5)^2$$
Take square roots:
$$x - 5 = \pm\, 2$$
$x = 5 + 2 = 7$
$x = 5 - 2 = 3$
The $x$-intercepts are $(3,\; 0)$ and $(7,\; 0)$.
Both are exactly $2$ units from $x = 5$, which matches the mental check.
9. Common Mistakes
Sign mistake in the vertex
In $y = (x+2)^2 - 3$, the vertex is $(-2,\; -3)$, not $(2,\; -3)$.
Confusing the vertex with the $y$-intercept
The vertex happens when $x = h$, not necessarily when $x = 0$.
Forgetting symmetry
Points equally spaced from the axis should match in $y$-value.
Dropping the square
In $(x-4)^2$, the whole binomial is squared — not just the $4$.
Ignoring the coefficient $a$
That number changes both the direction and the shape of the graph.
10. Rewriting into Vertex Form
Sometimes a quadratic starts in standard form:
$$y = ax^2 + bx + c$$
To find the vertex more easily, rewrite it in vertex form by completing the square.
Rewrite
$$y = x^2 + 8x + 11$$
in vertex form, and state the vertex.
Solution
Group the first two terms:
$$y = (x^2 + 8x) + 11$$
Take half of $8$, which is $4$, and square it: $4^2 = 16$.
Add and subtract $16$:
$$y = (x^2 + 8x + 16) - 16 + 11$$
Factor the perfect square trinomial:
$$y = (x+4)^2 - 5$$
Vertex form: $y = (x+4)^2 - 5$
Vertex: $(-4,\; -5)$
11. Real-World Model Connection
Quadratic functions often model a situation that rises to a peak and then falls, or falls to a lowest point and then rises.
Common examples:
- • height of a thrown object
- • area problems
- • profit or revenue models
- • motion over time
The vertex usually represents the most important moment in the situation — the peak height, the minimum cost, the maximum revenue, etc.
A fountain sends water into the air. Its height is modeled by
$$h(t) = -4(t-3)^2 + 36$$
where $h$ is the height in feet and $t$ is time in seconds. Find the maximum height, when the water reaches it, and when the water is at ground level.
Solution
The equation is already in vertex form. The vertex is $(3,\; 36)$.
Maximum height
$36$ feet
Time of maximum
$t = 3$ seconds
Find when the water is at ground level — set $h(t) = 0$:
$$0 = -4(t-3)^2 + 36$$
$$-36 = -4(t-3)^2$$
$$9 = (t-3)^2$$
$$t - 3 = \pm\, 3$$
$$t = 6 \quad \text{or} \quad t = 0$$
The water is at ground level at $t = 0$ (the start) and again at $t = 6$ seconds.
12. Final Summary
A quadratic in vertex form looks like $y = a(x-h)^2 + k$
The vertex is $(h,\; k)$ and the axis of symmetry is $x = h$
The sign of $a$ tells whether the parabola opens up or down
The size of $|a|$ tells whether the graph is wider or narrower than $y = x^2$
The vertex gives the maximum or minimum value of the function
You can write a quadratic from a vertex and one point by solving for $a$
You can find intercepts by setting $y = 0$ and solving
You can rewrite standard form into vertex form by completing the square
In real situations, the vertex often represents the highest or lowest value in the model